hcju/proofwikiqa huggingface dataset:
From Convex Real Function is Left-Hand and Right-Hand Differentiable, we have: :$\displaystyle \lim_{h \to 0^-} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^-} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^-} h}\right) = 0$ and similarly: :$\displaystyle \lim_{h \to 0^+} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^+} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^+} h}\right) = 0$ {{qed}}
Let $f$ be convex on $\left({a \,.\,.\, b}\right)$. Then by definition of convexity: :$\forall x_1, x_2, x_3 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$ Let $0 < h_1 < h_2$. Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then: :$\dfrac {f \left({x + h_1}\right) - f \left({x}\right)} {h_1} \le \dfrac {f \left({x + h_2}\right) - f \left({x}\right)} {h_2}$ Hence the function $F \left({h}\right) = \dfrac {f \left({x + h}\right) - f \left({x}\right)} h$ increases in some $\left({0 \,.\,.\, \delta}\right)$. Thus from Limit of Increasing Function it follows that both $\displaystyle \lim_{h \to 0^+} F \left({h}\right) = f'_+ \left({x}\right)$ and $\displaystyle \lim_{h \to 0^-} F \left({h}\right) = f'_-\left({x}\right)$ exist. {{qed}}
=== Necessary Condition === Let $f$ be convex on $\openint a b$. Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that: :$x_1 < x_2 < x_3 < x_4$ By the definition of convex function: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \le \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$ Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$. Thus: :$\map {f'} {x_1} \le \map {f'} {x_4}$ Hence $f'$ is increasing on $\openint a b$. {{qed|lemma}} === Sufficient Condition === Let $f'$ be increasing on $\openint a b$. Let $x_1, x_2, x_3 \in \openint a b$ such that $x_1 < x_2 < x_3$. By the Mean Value Theorem: :$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$ :$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$ where $x_1 < \xi < x_2 < \eta < x_3$. Since $f'$ is increasing: :$\map {f'} \xi \le \map {f'} \eta$ Thus: :$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ Hence $f$ is convex by definition. {{qed}}
Let: :$X = f \left({x}\right) \in J$ :$Y = f \left({y}\right) \in J$. From the definition of convex: : $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$ Let $f$ be strictly increasing on $I$. From Inverse of Strictly Monotone Function it follows that, $f^{-1}$ is strictly increasing on $J$. Thus: : $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \le f^{-1} \left({\alpha X + \beta Y}\right)$ Hence $f^{-1}$ is concave on $J$. {{qed}}
This follows directly from the definition of definite integral: From Continuous Image of Closed Interval is Closed Interval it follows that $m$ and $M$ both exist. The closed interval $\closedint a b$ is a finite subdivision of itself. By definition, the upper sum is $M \paren {b - a}$, and the lower sum is $m \paren {b - a}$. The result follows. {{qed}}
If $p = 2$ the result is obvious. Therefore we assume that $p$ is an odd prime. === Necessary Condition === {{:Wilson's Theorem/Necessary Condition}} === Sufficient Condition === {{:Wilson's Theorem/Sufficient Condition}}
Let $F$ be a primitive of $f$ on $\closedint a b$. By Primitive of Constant Multiple of Function, $H = c F$ is a primitive of $c f$ on $\closedint a b$. Hence by the Fundamental Theorem of Calculus: {{begin-eqn}} {{eqn | l = \int_a^b c \map f x \rd x       | r = \bigintlimits {c \map F x} a b       | c =  }} {{eqn | r = c \bigintlimits {\map F x} a b       | c =  }} {{eqn | r = c \int_a^b \map f x \rd x       | c =  }} {{end-eqn}} {{qed}}
Let $F$ be a primitive of $f$ on $\closedint a b$. By Linear Combination of Definite Integrals: :$\displaystyle \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$ for real functions $f$ and $g$ which are integrable on the closed interval $\closedint a b$, where $\lambda$ and $\mu$ be real numbers. The result follows by setting $\lambda = c$ and $\mu = 0$. {{qed}}
By Continuous Real Function is Darboux Integrable, $f$ is integrable on $I$. The result follows by application of Sum of Integrals on Adjacent Intervals for Integrable Functions. {{qed}}
=== Necessary Condition === Suppose $G$ is a primitive for $f$. Then $F - G$ is continuous on $\closedint a b$, differentiable on $\openint a b$, and for any $x \in \openint a b$, we have: {{begin-eqn}} {{eqn | l = \map {D_x} {\map F x - \map G x}       | r = \map {D_x} {\map F x} - \map {D_x} {\map G x}       | c = Sum Rule for Derivatives }} {{eqn | r = \map f x - \map f x       | c = $F, G$ are a primitives for $f$ }} {{eqn | r = 0 }} {{end-eqn}} From Zero Derivative implies Constant Function it follows that $F - G$ is constant on $\closedint a b$, hence the result. {{qed|lemma}} === Sufficient Condition === Now suppose $\map G x = \map F x + c$. We compute: {{begin-eqn}} {{eqn | l = D_x \map G x       | r = \map {D_x} {\map F x + c} }} {{eqn | r = \map {D_x} {\map F x} + 0       | c = Sum Rule for Derivatives and Derivative of Constant }} {{eqn | r = \map f x       | c = $F$ is a primitive for $f$ }} {{end-eqn}} Hence $G$ is also a primitive for $f$. {{qed}}
Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a finite subdivision of $\closedint a b$. Let $\map {L^{\paren {f + c} } } P$ be the lower sum of $\map f x + c$ on $\closedint a b$ belonging to $P$. Let: :$\ds m_k^{\paren {f + c} } = \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x + c}$ where $k \in \set {0, 1, \ldots, n}$. So: {{begin-eqn}} {{eqn | l = m_k^{\paren {f + c} }       | r = \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x + c}       | c =  }} {{eqn | r = c + \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x}       | c = }} {{eqn | r = c + m_k^{\paren f}       | c = }} {{end-eqn}} It follows that: {{begin-eqn}} {{eqn | l = \map {L^{\paren {f + c} } } P       | r = \sum_{k \mathop = 1}^n {m_k^{\paren {f + c} } } \paren {x_k - x_{k - 1} }       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^n m_k^{\paren f} \paren {x_k - x_{k - 1} } + c \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }       | c =  }} {{eqn | r = \map {L^{\paren f} } P + c \paren {b - a}       | c = as $\ds \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }$ telescopes }} {{end-eqn}} So from the definition of definite integral, it follows that: {{begin-eqn}} {{eqn | l = \int_a^b \paren {\map f t + c} \rd t       | r = \map {\sup_P} {\map {L^{\paren {f + c} } } P}       | c =  }} {{eqn | r = \map {\sup_P} {\map {L^{\paren f} } P + c \paren {b - a} }       | c =  }} {{eqn | r = \map {\sup_P} {\map {L^{\paren f} } P} + c \paren {b - a}       | c =  }} {{eqn | r = \int_a^b \map f t \rd t + c \paren {b - a}       | c =  }} {{end-eqn}} {{Qed}}
We have the result $f$ is bounded by Continuous Real Function is Bounded. By Condition for Darboux Integrability, it suffices to show that for all $\epsilon > 0$, there exists a subdivision $P$ of $\closedint a b$ such that: :$\map U P – \map L P < \epsilon$ where $\map U P$ and $\map L P$ denote the upper sum and lower sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$. Let $\epsilon > 0$. We have the result Continuous Function on Closed Interval is Uniformly Continuous. By the definition of uniform continuity, there exists a $\delta > 0$ such that if $x, y \in \closedint a b$ are such that $\size {x – y} < \delta$, then: :$\size {\map f x – \map f y} < \dfrac \epsilon {b - a}$ Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a subdivision of $\closedint a b$ such that: :$\displaystyle \max_{1 \mathop \le k \mathop \le n} \paren {x_k – x_{k - 1} } < \delta$ For all integers $k$ satisfying $1 \le k \le n$, it follows from the Heine-Borel theorem that $\closedint {x_{k - 1} } {x_k}$ is compact. So we can apply Corollary 3 to Continuous Image of Compact Space is Compact to conclude that there exist $u_k, v_k \in \closedint {x_{k - 1} } {x_k}$ such that: {{begin-eqn}} {{eqn | l = \map f {u_k}       | r = \sup \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} } }} {{eqn | l = \map f {v_k}       | r = \inf \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} } }} {{end-eqn}} By assumption, $x_k – x_{k - 1} < \delta$, so: :$\size {u_k – v_k} < \delta$ It follows from the definition of $\delta$ that: :$\displaystyle \map f {u_k} – \map f {v_k} < \frac \epsilon {b – a}$ This gives: {{begin-eqn}} {{eqn | l = \map U P – \map L P       | r = \sum_{k \mathop = 1}^n \map f {u_k} \paren {x_k – x_{k - 1} } - \sum_{k \mathop = 1}^n \map f {v_k} \paren {x_k – x_{k - 1} } }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\map f {u_k} – \map f {v_k} } \paren {x_k – x_{k - 1} } }} {{eqn | o = <       | r = \frac \epsilon {b – a} \sum_{k \mathop = 1}^n \paren {x_k – x_{k - 1} } }} {{eqn | r = \frac \epsilon {b – a} \paren {x_n – x_0} }} {{eqn | r = \epsilon }} {{end-eqn}} as desired. {{qed}}
Follows directly from the definition of definite integral. There is only one finite subdivision of $\closedint c c$ and that is $\set c$. Both the lower sum and upper sum of $\map f x$ on $\closedint c c$ belonging to the finite subdivision $\set c$ are equal to zero. Hence the result. {{qed}}
Suppose that $\forall t \in \closedint a b: \map f t \le \map g t$. From the Fundamental Theorem of Calculus, $g - f$ has a primitive on $\closedint a b$. Let $H$ be such a primitive. Then: :$\forall t \in \closedint a b: D_t \map H t = \map g t - \map f t \ge 0$ By Derivative of Monotone Function, it follows that $H$ is increasing on $\closedint a b$. Thus: :$\map H b \ge \map H a$ Hence: :$\displaystyle \int_a^b \map g t - \map f t \rd t = \map H b - \map H a \ge 0$ The proof for the second case is similar. {{qed}}
From Negative of Absolute Value, we have for all $a \in \closedint a b$: :$-\size {\map f t} \le \map f t \le \size {\map f t}$ Thus from Relative Sizes of Definite Integrals: :$\displaystyle -\int_a^b \size {\map f t} \rd t \le \int_a^b \map f t \rd t \le \int_a^b \size {\map f t} \rd t$ Hence the result. {{qed}}
From Upper and Lower Bounds of Integral, we have that: :$\displaystyle m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$ where: :$M$ is the maximum and: :$m$ is the minimum of $\map f x$ on $\closedint a b$. Since $f$ decreases, $M = \map f a$ and $m = \map f b$. Thus it follows that: :$\displaystyle \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$ as $\paren {k + 1} - k = 1$. Thus: {{begin-eqn}} {{eqn | l = \Delta_{n + 1} - \Delta_n       | r = \paren {\sum_{k \mathop = 1}^{n + 1} \map f k - \int_1^{n + 1} \map f x \rd x} - \paren {\sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x}       | c =  }} {{eqn | r = \map f {n + 1} - \int_n^{n + 1} \map f x \rd x       | c =  }} {{eqn | o = \le       | r = \map f {n + 1} - \map f {n + 1}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Thus $\sequence {\Delta_n}$ is decreasing. {{qed|lemma}} Also: {{begin-eqn}} {{eqn | l = \Delta_n       | r = \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \int_k^{k + 1} \map f x \rd x       | c =  }} {{eqn | o = \ge       | r = \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \map f k       | c =  }} {{eqn | r = \map f n       | c =  }} {{eqn | o = \ge       | r = 0       | c =  }} {{end-eqn}} {{qed|lemma}} Hence the result. {{qed}}
We construct such a set. For $x, y \in \left[{0 \,.\,.\, 1}\right)$, define the sum modulo 1: :$x +_1 y = \begin{cases} x + y & : x + y < 1 \\ x + y - 1 & : x + y \ge 1 \end{cases}$ Let $E \subset \left[{0 \,.\,.\, 1}\right)$ be a measurable set. Let $E_1 = E \cap \left[{0 \,.\,.\, 1 - x}\right)$ and $E_2 = E \cap \left[{1 - x \,.\,.\, 1}\right)$. By Measure of Interval is Length, these disjoint intervals are measurable. By Measurable Sets form Algebra of Sets, so are these intersections $E_1$ and $E_2$. So: :$m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$ We have: : $E_1 +_1 x = E_1 + x$ By Lebesgue Measure is Translation-Invariant: : $m \left({E_1 +_1 x}\right) = m \left({E_1}\right)$ Also: : $E_2 +_1 x = E_2 + x - 1$, and so $m \left({E_2 +_1 x}\right) = m \left({E_2}\right)$ Then we have: : $m \left({E +_1 x}\right) = m \left({E_1 +_1 x}\right) + m \left({E_2 +_1 x}\right) = m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$ So, for each $x \in \left[{0 \,.\,.\, 1}\right)$, the set $E +_1 x$ is measurable and: :$m \left({E + x}\right) = m \left({E}\right)$ Taking, as before, $x, y \in \left[{0 \,.\,.\, 1}\right)$, define the relation: :$x \sim y \iff x - y \in \Q$ where $\Q$ is the set of rational numbers. By Difference is Rational is Equivalence Relation, $\sim$ is an equivalence relation. As this is an equivalence relation we can invoke the fundamental theorem on equivalence relations. Hence $\sim$ partitions $\left[{0 \,.\,.\, 1}\right)$ into equivalence classes. By the axiom of choice, there is a set $P$ which contains exactly one element from each equivalence class. Let $\left\{{r_i}\right\}_{i \mathop = 0}^\infty$ be an enumeration of the rational numbers in $\left[{0 \,.\,.\, 1}\right)$ with $r_0 = 0$. Let $P_i := P +_1 r_i$. Then $P_0 = P$. Let $x \in P_i \cap P_j$. Then: :$x = p_i + r_i = p_j + r_j$ where $p_i, p_j$ are elements of $P$. But then $p_i - p_j$ is a rational number. Since $P$ has only one element from each equivalence class: :$i = j$ The $P_i$ are pairwise disjoint. Each real number $x \in \left[{0 \,.\,.\, 1}\right)$ is in ''some'' equivalence class and hence is equivalent to an element of $P$. But if $x$ differs from an element in $P$ by the rational number $r_i$, then $x \in P_i$ and so: :$\displaystyle \bigcup P_i = \left[{0 \,.\,.\, 1}\right)$ Since each $P_i$ is a translation modulo $1$ of $P$, each $P_i$ will be measurable if $P$ is, with measure $m \left({P_i}\right) = m \left({P}\right)$. But if this were the case, then: :$\displaystyle m \left[{0 \,.\,.\, 1}\right) = \sum_{i \mathop = 1}^\infty m \left({P_i}\right) = \sum_{i \mathop = 1}^\infty m \left({P}\right)$ Therefore: : $m \left({P}\right) = 0$ implies $m \left[{0 \,.\,.\, 1}\right) = 0$ and: : $m \left({P}\right) \ne 0$ implies $m \left[{0 \,.\,.\, 1}\right) = \infty$ This contradicts Measure of Interval is Length. So the set $P$ is not measurable. {{Qed}} {{AoC||4}} {{BPI}} {{explain|While BPI has been invoked as being necessary for this theorem, no explanation has been added as to why, or how, or where it would be applied.}}
:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$  where $\displaystyle s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$ From Ordering of Reciprocals: : $\forall m, n \in \N_{>0}: m < n: \dfrac 1 m > \dfrac 1 n$ so each of the summands in a given $s_k$ is greater than $\dfrac 1 {2^{k + 1} }$. The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so: :$s_k > \dfrac{2^k} {2^{k + 1} } = \dfrac 1 2$ Hence the harmonic sum $H_{2^m}$ satisfies the following inequality: {{begin-eqn}} {{eqn | l = H_{2^m}       | r = \sum_{n \mathop = 1}^{2^m} \frac 1 n       | c =  }} {{eqn | o = >       | r = \sum_{n \mathop = 1}^{2^m - 1} \frac 1 n       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^m \left({s_k}\right)       | c =  }} {{eqn | o = >       | r = 1 + \sum_{a \mathop = 0}^m \frac 1 2       | c =  }} {{eqn | r = 1 + \frac m 2       | c =  }} {{end-eqn}} The {{RHS}} diverges, from the $n$th term test. The result follows from the the Comparison Test for Divergence. {{qed}}
Observe that all the terms of the harmonic series are strictly positive. From Reciprocal Sequence is Strictly Decreasing, the terms are decreasing. Hence the Cauchy Condensation Test can be applied, and we examine the convergence of: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty 2^n \frac 1 {2^n}       | r = \sum_{n \mathop = 1}^\infty 1 }} {{end-eqn}} This diverges, from the $n$th term test. Hence $\displaystyle \sum \frac 1 n$ also diverges. {{qed}}
We have that the Integral of Reciprocal is Divergent. Hence from the Integral Test, the harmonic series also diverges. {{qed}}
For all $N \in \N$: :$\dfrac 1 N + \dfrac 1 {N + 1} + \cdots + \dfrac 1 {2 N} > N \cdot \dfrac 1 {2 N} = \dfrac 1 2$ Hence, by Cauchy's Convergence Criterion for Series, the Harmonic series is divergent. {{qed}}
Assume that for $G \ge 4$ that $\displaystyle \sum_{n = G}^\infty \frac{1}{n} = L < \infty$. Namely, that for some number $G$, there is a tail of the harmonic series which converges.  Then from Definition:Series/Sequence of Partial Sums: $s_N := \sum_{n = G}^N \frac{1}{n}$ is the partial sum of the above series. Which yields the sequence $\{ s_N \}$ of partial sums.   And, from Definition:Convergent Series we have that $\displaystyle \sum_{n = G}^\infty \frac{1}{n}$ converges iff $\{ s_N\}$ converges.  From Constant Sequence Converges to Constant in Normed Division Ring: The constant sequence $\{ G \}$ has limit $G$.  Note: $\R$ is a normed division ring as it is a field.  By Combination Theorem for Sequences/Real/Product Rule: The product of the sequences $\{ G \}$ and $\{ s_N\}$ has limit $GL$. Namely, the sequence $\{Gs_N\}$ has limit $GL$, by the opening assumption. $GL = \displaystyle \sum_{n = G}^\infty \frac{G}{n} =  \underbrace {1}_{s_0} + \underbrace{ \frac{G}{G+1} + \ldots + \frac{G}{G+4}}_{s_1} + \underbrace { \frac{G}{G+5} + \ldots + \frac{G}{G+12} }_{s_2} + \underbrace { \frac{G}{G+13} + \ldots + \frac{G}{G+28} }_{s_3} + \ldots $  Where $s_0 = 1, s_1 = \frac{G}{G+1} + \ldots + \frac{G}{G+4}$ and for $k \ge 2, s_k = \displaystyle \sum_{i =  2^k + 2^{k-1} + \ldots 2^2 + 1}^{2^{k+1} + 2^{k} + \ldots 2^2} \frac{G}{G+i} $. From the above, $s_0 = 1, s_1 \ge \frac{4G}{G+4} \ge 1$ by inspection.  And for $k \ge 2$ then $s_k \ge \frac{2^{k+1}G}{G+2^{k+2}}$ since $\frac{G}{G+2^{k+2}}$ is smaller than the smallest summand of $s_k$. If summed $2^{k+1}$ many times, $2^{k+1}$ being the number of summands in $s_k$, it yields a result less than $s_k$. Note: The smallest summand of $s_k$ is $\frac{G}{G+2^{k+1}+ \ldots + 2^2}$. Claim:  For $k \ge 1, G \ge 4$ We have $\frac{2^{k+1}G}{G+2^{k+2}} \ge 1$  Proof:  $\frac{2^{k+1}G}{G+2^{k+2}} \ge 1 \iff k + 1 \ge \log_2\frac{G}{G-2}$. If $G = 4$ then the righthand side of the second inequality is $\log_2(2) = 1$. If $G > 4$, then $ 1 < \frac{G}{G-2} < 2$. Namely as $G \uparrow$ we have $\frac{G}{G-2} \to 1$ meaning $\log_2\frac{G}{G-2} \to 0$. {{qed|lemma}} Now, $\displaystyle GL = s_0 + s_1 + s_2 + s_3 + \ldots \ge 1 + \frac{4G}{G+4} + \frac{8G}{G+16 } + \frac{16G}{G+32} + \ldots  \ge 1 + 1 + 1 + 1 +  \ldots \to \infty > GL$ Deriving a contradiction.  Hence , the series does not converge which implies the sequence $\{s_N\}$ does not converge. Therefore by  Tail of Convergent Sequence  : A sequence $a_n$ converges iff the sequence $a_{n+N}, N \in \N$ converges. We have the tail of the harmonic series diverges for any $G$ thus the harmonic series will diverge. {{qed}}
By Sum of Reciprocals of Primes is Divergent: Lemma: :$\displaystyle \lim_{n \mathop \to \infty} \paren {\map \ln {\map \ln n} - \frac 1 2} = +\infty$ {{qed|lemma}} It remains to be proved that: :$\displaystyle \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - \frac 1 2$ Assume all sums and product over $p$ are over the set of prime numbers. Let $n \ge 1$. {{begin-eqn}} {{eqn | l = \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1}       | r = \prod_{p \mathop \le n} \sum_{k \mathop = 0}^\infty \frac 1 {p^k}       | c = Sum of Infinite Geometric Sequence }} {{eqn | o = \ge       | r = \sum_{k \mathop = 1}^n \frac 1 k        | c = Fundamental Theorem of Arithmetic }} {{eqn | o = >       | r = \int_1^n \frac 1 x \rd x       | c = Integral Test }} {{eqn | n = 1       | r = \ln n       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \ln \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1}       | r = \sum_{p \mathop \le n} \map \ln {1 - \frac 1 p}^{-1}       | c = Sum of Logarithms }} {{eqn | r = \sum_{p \mathop \le n} \sum_{k \mathop = 1}^\infty \frac 1 {k p^k}       | c = Power Series Expansion for $\map \ln {1 + x}$ }} {{eqn | o = <       | r = \sum_{p \mathop \le n} \frac 1 p + \sum_{p \mathop \le n} \frac 1 {2 p^2} \paren {\sum_{k \mathop = 0}^\infty \frac 1 {p^k} }       | c =  }} {{eqn | r = \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{p \mathop \le n} \frac 1 {p \paren {p - 1} }       | c = Sum of Infinite Geometric Sequence }} {{eqn | o = <       | r = \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{n \mathop = 2}^\infty \frac 1 {n \paren {n - 1} }       | c =  }} {{eqn | r = \sum_{p \mathop \le n} \frac 1 p + \frac 1 2       | c = {{Defof|Telescoping Series}} }} {{eqn | o = >       | l = \sum_{p \mathop \le n} \frac 1 p       | r = \ln \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1} - \frac 1 2       | c =  }} {{eqn | o = >       | r = \map \ln {\ln n} - \frac 1 2       | c = by $(1)$ }} {{end-eqn}} {{qed}}
Let $n \in \N$ be a natural number. Let $p_n$ denote the $n$th prime number. Consider the product: :$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$ By Sum of Geometric Sequence, we have: {{begin-eqn}} {{eqn | l = \frac 1 {1 - \frac 1 2}       | r = 1 + \frac 1 2 + \frac 1 {2^2} + \cdots       | c =  }} {{eqn | l = \frac 1 {1 - \frac 1 3}       | r = 1 + \frac 1 3 + \frac 1 {3^2} + \cdots       | c =  }} {{eqn | l = \frac 1 {1 - \frac 1 5}       | r = 1 + \frac 1 5 + \frac 1 {5^2} + \cdots       | c =  }} {{eqn | o = \cdots       | c =  }} {{eqn | l = \frac 1 {1 - \frac 1 {p_n} }       | r = 1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots       | c =  }} {{end-eqn}} Consider what happens when all these series are multiplied together. A new series will be generated whose terms consist of all possible products of one term selected from each of the series on the {{RHS}}. This new series will converge in any order to the product of the terms on the {{LHS}}. By the Fundamental Theorem of Arithmetic, every integer greater than $1$ is uniquely expressible as a product of powers of different primes. Hence the product of these series is the series of reciprocals of all (strictly) positive integers whose prime factors are no greater than $p_n$. In particular, all (strictly) positive integers up to $p_n$ have this property. So: :$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$ {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}       | o = \ge       | r = \sum_{k \mathop = 1}^{p_n} \frac 1 k       | c =  }} {{eqn | o = >       | r = \int_1^{p_n + 1} \dfrac {\mathrm d x} x       | c =  }} {{eqn | r = \map \ln {p_n + 1}       | c =  }} {{eqn | r = \ln p_n       | c =  }} {{end-eqn}} It follows by taking reciprocals that: :$\displaystyle \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$ Taking logarithms of each side: :$(1): \quad \displaystyle \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$ Next, note that the line $y = 2 x$ in the cartesian plane lies below the curve $y = \map \ln {1 + x}$ on the interval $\closedint {-\frac 1 2} 0$. Also note that all primes are greater than or equal to $2$. Thus it follows that: :$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$ Applying this to $(1)$ yields: :$\displaystyle -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$ and so: :$\displaystyle \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$ But: :$\displaystyle \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$ and so the series: :$\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$ is divergent. {{qed}}