hcju/proofwikiqa huggingface dataset:
We observe that: {{begin-eqn}} {{eqn | l = \frac 1 j - \frac 1 {j + 2}       | r = \frac {\left({j + 2}\right) - j} {j \left({j + 2}\right)}       | c =  }} {{eqn | r = \frac 2 {\left({j \left({j + 2}\right)}\right)}       | c =  }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^n \frac 1 {j \left({j + 2}\right)}       | r = \frac 1 2 \sum_{j \mathop = 1}^n \left({\frac 1 j - \frac 1 {j + 2} }\right)       | c =  }} {{eqn | r = \frac 1 2 \left({\left({\frac 1 1 - \frac 1 3}\right) + \left({\frac 1 2 - \frac 1 4}\right) + \left({\frac 1 3 - \frac 1 5}\right) + \cdots + \left({\frac 1 {n - 1} - \frac 1 {n + 1} }\right) + \left({\frac 1 n - \frac 1 {n + 2} }\right)}\right)       | c =  }} {{eqn | r = \frac 1 2 \left({1 + \frac 1 2 - \frac 1 {n + 1} - \frac 1 {n + 2} }\right)       | c = Telescoping Series }} {{eqn | r = \frac 3 4 - \frac {\left({n + 2}\right) + \left({n + 1}\right)} {2 \left({n + 1}\right) \left({n + 2}\right)}       | c =  }} {{eqn | r = \frac 3 4 - \frac {2 n + 3} {2 \left({n + 1}\right) \left({n + 2}\right)}       | c =  }} {{end-eqn}} {{qed}} Category:Sums of Sequences Category:Reciprocals od0yoj6ihkbagwbz6odwmwc4tyb6knk
Note that we have:  :$\displaystyle \frac 1 {2 j + 1} - \frac 1 {2 j + 3} = \frac 2 {\paren {2 j + 1} \paren {2 j + 3} }$ So: {{begin-eqn}} {{eqn	| l = \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1}^2 \paren {2 j + 3}^2} 	| r = \frac 1 4 \sum_{j \mathop = 0}^\infty \paren {\frac 1 {2 j + 1} - \frac 1 {2 j + 3} }^2 }} {{eqn	| r = \frac 1 4 \sum_{j \mathop = 0}^\infty \paren {\frac 1 {\paren {2 j + 1}^2} - \frac 2 {\paren {2 j + 1} \paren {2 j + 3} } + \frac 1 {\paren {2 j + 3}^2} } 	| c = Square of Sum }} {{eqn	| r = \frac 1 4 \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1}^2} - \frac 1 2 \sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1} \paren {2 j + 3} } + \frac 1 4 \paren {\sum_{j \mathop = 0}^\infty \frac 1 {\paren {2 j + 1}^2} - 1} }} {{eqn	| r = \frac {\pi^2} {32} - \frac 1 4 + \frac {\pi^2} {32} - \frac 1 4 	| c = Sum of Reciprocals of Squares of Odd Integers, Sum of Sequence of Products of Consecutive Odd Reciprocals }} {{eqn	| r = \frac {\pi^2} {16} - \frac 1 2 }} {{eqn	| r = \frac {\pi^2 - 8} {16} }} {{end-eqn}} {{qed}}
We use partial fraction expansion to expand $\dfrac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}$. Let $\dfrac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2} = \dfrac A j + \dfrac B {j^2} + \dfrac C {j + 1} + \dfrac D {\paren {j + 1}^2} + \dfrac E {j + 2} + \dfrac F {\paren {j + 2}^2}$. This has solutions $A = - \dfrac 3 4, B = \dfrac 1 4, C = 0, D = 1, E = \dfrac 3 4, F = \dfrac 1 4$. Thus: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j^2 \paren {j + 1}^2 \paren {j + 2}^2}       | r = \sum_{j \mathop = 1}^\infty \paren {-\frac 3 {4j} + \frac 1 {4 j^2} + \frac 1 {\paren {j + 1}^2} + \frac 3 {4 \paren {j + 2} } + \frac 1 {4 \paren {j + 2}^2} } }} {{eqn | r = \sum_{j \mathop = 1}^\infty \paren {\frac 1 {4 j^2} + \frac 1 {\paren {j + 1}^2} + \frac {3 j - 3 \paren {j + 2} } {4 j \paren {j + 2} } + \frac 1 {4 \paren {j + 2}^2} } }} {{eqn | r = \sum_{j \mathop = 1}^\infty \frac 1 {4 j^2} + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {j + 1}^2} + \sum_{j \mathop = 1}^\infty \frac 1 {4 \paren {j + 2}^2} - \sum_{j \mathop = 1}^\infty \frac 6 {4 j \paren {j + 2} } }} {{eqn | r = \frac 1 4 \sum_{j \mathop = 1}^\infty \frac 1 {j^2} + \sum_{j \mathop = 2}^\infty \frac 1 {j^2} + \frac 1 4 \sum_{j \mathop = 3}^\infty \frac 1 {j^2} - \frac 3 2 \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 2} }       | c = Translation of Index Variable of Summation }} {{eqn | r = \frac 3 2 \sum_{j \mathop = 1}^\infty \frac 1 {j^2} - 1 - \frac 1 4 - \frac 1 {16} - \frac 3 2 \paren {\frac 3 4}       | c = Corollary to Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals }} {{eqn | r = \frac 3 2 \paren {\frac {\pi^2} 6} - \frac {21} {16} - \frac 9 8       | c = Basel Problem }} {{eqn | r = \frac {4 \pi^2 - 39} {16} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^1 \frac {u^{a - 1} \rd u} {1 + u^d} 	| r = \int_0^1 \frac {u^{a - 1} \rd u} {1 - \paren {-u^d} } }} {{eqn	| r = \int_0^1 u^{a - 1} \sum_{j \mathop = 0}^\infty \paren {-1}^j u^{j d} \rd u 	| c = Sum of Infinite Geometric Sequence }} {{eqn	| r = \int_0^1 \sum_{j \mathop = 0}^\infty \paren {-1}^j u^{a - 1 + j d} \rd u }} {{eqn	| r = \sum_{j \mathop = 0}^\infty \paren {-1}^j \int_0^1 u^{a - 1 + j d} \rd u 	| c = Fubini's Theorem }} {{eqn	| r = \sum_{j \mathop = 0}^\infty \paren {-1}^j \intlimits {\frac {u^{a + j d} } {a + j d} } 0 1 	| c = Primitive of Power, Fundamental Theorem of Calculus }} {{eqn	| r = \sum_{j \mathop = 0}^\infty \frac {\paren {-1}^j} {a + j d} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }       | r = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j}^{2 n} } + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }       | c =  }} {{eqn | r = \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} } + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }       | c =  }} {{eqn | ll= \leadsto       | l = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}       | r = \frac 1 {2^{2 n} } \times \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} + \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }       | c = Riemann Zeta Function at Even Integers }} {{eqn | ll= \leadsto       | l = \sum_{j \mathop = 1}^\infty \frac 1 {\paren {2 j - 1}^{2 n} }       | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} - \frac 1 {2^{2 n} } \times \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} }}  {{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n} \pi^{2 n} } {2 \paren {2 n}!} - \paren {-1}^{n + 1} \dfrac {B_{2 n} \pi^{2 n} } {2 \paren {2 n}!} }} {{eqn | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n} - 1} \pi^{2 n} } {2 \paren {2 n}!} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \left({-1}\right)^{j + 1} \frac 1 {j^{2 n} }       | r = \sum_{j \mathop = 1}^\infty \frac 1 {\left({2 j - 1}\right)^{2 n} } - \sum_{j \mathop = 1}^\infty \frac 1 {\left({2 j}\right)^{2 n} }       | c = separating odd positive terms from even negative terms }} {{eqn | r = \sum_{j \mathop = 1}^\infty \frac 1 {\left({2 j - 1}\right)^{2 n} } - \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} } }} {{eqn | r = \left({-1}\right)^{n + 1} \dfrac {B_{2 n} \left({2^{2 n} - 1}\right) \pi^{2 n} } {2 \left({2 n}\right)!} - \frac 1 {2^{2 n} } \sum_{j \mathop = 1}^\infty \frac 1 {j^{2 n} }       | c = Sum of Reciprocals of Even Powers of Odd Integers }} {{eqn | r = \left({-1}\right)^{n + 1} \dfrac {B_{2 n} \left({2^{2 n} - 1}\right) \pi^{2 n} } {2 \left({2 n}\right)!} - \frac 1 {2^{2 n} } \left({-1}\right)^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\left({2 n}\right)!}       | c = Riemann Zeta Function at Even Integers }} {{eqn | r = \left({-1}\right)^{n + 1} \dfrac {2^{2 n - 1} B_{2 n} \left({2^{2 n} - 1}\right) \pi^{2 n} - B_{2 n} 2^{2 n - 1} \pi^{2 n} } {2^{2 n} \left({2 n}\right)!} }} {{eqn | r = \left({-1}\right)^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \left({2^{2 n} - 2}\right) \pi^{2 n} } {2^{2 n} \left({2 n}\right)!} }} {{eqn | r = \left({-1}\right)^{n + 1} \dfrac {B_{2 n} \left({2^{2 n - 1} - 1}\right) \pi^{2 n} } {\left({2 n}\right)!} }} {{end-eqn}} {{qed}}
From Euler's Formula: :$e^{i \theta} = \cos \theta + i \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^\infty r^k \cos k x       | r = \map \Re {\sum_{k \mathop = 0}^\infty r^k e^{i k x} }       | c =  }} {{eqn | r = \map \Re {\sum_{k \mathop = 0}^\infty \paren {r e^{i x} }^k}       | c =  }} {{eqn | r = \map \Re {\frac 1 {1 - r e^{i x} } }       | c = Sum of Infinite Geometric Sequence: valid because $\size r < 1$ }} {{eqn | r = \map \Re {\frac {1 - r e^{-i x} } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }       | c =  }} {{eqn | r = \map \Re {\frac {1 - r e^{-i x} } {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }       | c =  }} {{eqn | r = \map \Re {\frac {1 - r \paren {\cos x - i \sin x} } {1 - 2 r \cos x + a^2} }       | c = Corollary to Euler's Formula }} {{eqn | r = \dfrac {1 - r \cos x} {1 - 2 r \cos x + r^2}        | c = after simplification }} {{end-eqn}} It is noted that when $x$ is a multiple of $2 \pi$ then: :$1 - 2 r \cos x + r^2 = 1 - 2 + 1 = 0$ leaving the {{RHS}} undefined. {{qed}}
From Euler's Formula: :$e^{i \theta} = \cos \theta + i \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n r^k \map \cos {k x}       | r = \map \Re {\sum_{k \mathop = 0}^n r^k e^{i k x} }       | c =  }} {{eqn | r = \map \Re {\sum_{k \mathop = 0}^n \paren {r e^{i x} }^k}       | c =  }} {{eqn | r = \map \Re {\frac {1 - r^{n + 1} e^{i \paren {n + 1} x} } {1 - r e^{i x} } }       | c = Sum of Geometric Sequence }} {{eqn | r = \map \Re {\frac {\paren {1 - r^{n + 1} e^{i \paren {n + 1} x} } \paren {1 - r e^{-i x} } } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }        | c =  }} {{eqn | r = \map \Re {\frac {r^{n + 2} e^{i n x} - r^{n + 1} e^{i \paren {n + 1} x} - r e^{-i x} + 1} {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }        | c =  }} {{eqn | r = \map \Re {\frac {r^{n + 2} \paren {\cos n x + i \sin n x} - r^{n + 1} \paren {\map \cos {n + 1} x + i \map \sin {n + 1} x} - r \paren {\cos x - i \sin x} + 1} {1 - 2 r \cos x + a^2} }       | c = Euler's Formula and Corollary to Euler's Formula }} {{eqn | r = \dfrac {r^{n + 2} \cos n x - r^{n + 1} \map \cos {n + 1} x - r \cos x + 1} {1 - 2 r \cos x + r^2}       | c = after simplification }} {{end-eqn}} It is noted that when $x$ is a multiple of $2 \pi$ then: :$1 - 2 r \cos x + r^2 = 1 - 2 + 1 = 0$ leaving the {{RHS}} undefined. {{qed}}
From Euler's Formula: :$e^{i \theta} = \cos \theta + i \sin \theta$ Hence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n r^k \map \sin {k x}       | r = \map \Im {\sum_{k \mathop = 1}^n r^k e^{i k x} }       | c =  }} {{eqn | r = \map \Im {\sum_{k \mathop = 0}^n \paren {r e^{i x} }^n}       | c = as $\map \Im {e^{i \times 0 \times x} } = \map \Im 1 = 0$ }} {{eqn | r = \map \Im {\frac {1 - r^{n + 1} e^{i \paren {n + 1} x} } {1 - r e^{i x} } }        | c = Sum of Geometric Sequence }} {{eqn | r = \map \Im {\frac {\paren {1 - r^{n + 1} e^{i \paren {n + 1} x} } \paren {1 - r e^{-i x} } } {\paren {1 - r e^{-i x} } \paren {1 - r e^{i x} } } }        | c =  }} {{eqn | r = \map \Im {\frac {r^{n + 2} e^{i n x} - r^{n + 1} e^{i \paren {n + 1} x} - r e^{-i x} + 1} {1 - r \paren {e^{i x} + e^{- i x} } + r^2} }       | c =  }} {{eqn | r = \map \Im {\frac {r^{n + 2} \paren {\cos n x + i \sin n x} - r^{n + 1} \paren {\map \cos {n + 1} x + i \map \sin {n + 1} x} - r \paren {\cos x - i \sin x} + 1} {1 - 2 r \cos x + a^2} }        | c = Euler's Formula and Corollary to Euler's Formula }} {{eqn | r = \dfrac {r \sin x - r^{n + 1} \map \sin {n + 1} x + r^{n + 2} \sin n x} {1 - 2 r \cos x + r^2}       | c = after simplification }} {{end-eqn}} It is noted that when $x$ is a multiple of $2 \pi$ then: :$1 - 2 r \cos x + r^2 = 1 - 2 + 1 = 0$ leaving the {{RHS}} undefined. {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {1 + x}       | r = \frac 1 {1 - \paren {-x} }       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-x}^k       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {1 + x}       | r = \left({1 + x}\right)^{-1}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({-1}\right)^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({\left({-1}\right) - j}\right)} {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({-\left({j + 1}\right)}\right)} {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c = {{Defof|Factorial}} and simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 1}^k \left({- j}\right)} {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c = Translation of Index Variable of Product }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({-1}\right)^k \displaystyle \prod_{j \mathop = 1}^k j} {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k x^k       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {1 + x}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k x^k       | c = Power Series Expansion of $\dfrac 1 {1 + x}$ }} {{eqn | ll= \leadsto       | l = \frac \d {\d x} \frac 1 {1 + x}       | r = \frac \d {\d x} \sum_{k \mathop = 0}^\infty \left({-1}\right)^k x^k       | c =  }} {{eqn | ll= \leadsto       | l = -\frac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k k x^{k - 1}       | c = differentiating {{WRT|Differentiation}} $x$ }} {{eqn | ll= \leadsto       | l = \frac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^{k - 1} k x^{k - 1}       | c = taking one of the $-1$s out }} {{eqn | r = \sum_{k \mathop = 1}^\infty \left({-1}\right)^{k - 1} k x^{k - 1}       | c = the term in $k = 0$ vanishes }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) k x^k       | c = Translation of Index Variable of Product }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\left({1 + x}\right)}       | r = \left({1 + x}\right)^{-2}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({-2}\right)^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({\left({-2}\right) - j}\right)} {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({-\left({j + 2}\right)}\right)} {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c = {{Defof|Factorial}} and simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 1}^k \left({-\left({j + 1}\right)}\right)} {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c = Translation of Index Variable of Product }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({-1}\right)^k \displaystyle \prod_{j \mathop = 1}^k \left({j + 1}\right)} {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) x^k       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn	| l = \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x 	| r = \int_0^\infty \intlimits {\frac {\map f {x t} } x} {t = b} a \rd x }} {{eqn	| r = \int_0^\infty \int_b^a \map {f'} {x t} \rd t \rd x 	| c = Fundamental Theorem of Calculus }}  {{eqn	| r = \int_b^a \int_0^\infty \map {f'} {x t} \rd x \rd t 	| c = Fubini's Theorem }} {{eqn	| r = \int_b^a \intlimits {\frac {\map f {x t} } t} {x = 0} \infty \rd t 	| c = Fundamental Theorem of Calculus }} {{eqn	| r = \int_b^a \frac {\map f \infty - \map f 0} t \rd t }} {{eqn	| r = \paren {\map f \infty - \map f 0} \paren {\ln a -\ln b} 	| c = Primitive of Reciprocal, Fundamental Theorem of Calculus }} {{eqn	| r = \paren {\map f \infty - \map f 0} \ln \frac a b 	| c = Difference of Logarithms }} {{end-eqn}} {{qed}} {{Namedfor|Giuliano Frullani|cat = Frullani}}
{{begin-eqn}} {{eqn | l = \frac 1 {\paren {1 + x} }       | r = \paren {1 + x}^{-3}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-3}^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-3} - j} } {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 3} } } {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c = {{Defof|Factorial}} and simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 1}^k \paren {-\paren {j + 2} } } {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c = Translation of Index Variable of Product }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \displaystyle \prod_{j \mathop = 1}^k \paren {j + 2} } {\displaystyle \prod_{j \mathop = 1}^k j} x^k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {k + 2} \paren {k + 1} } 2 x^k       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) x^k       | c = Power Series Expansion of $\dfrac 1 {\left({1 + x}\right)^2}$ }} {{eqn | ll= \leadsto       | l = \frac \d {\d x} \frac 1 {\left({1 + x}\right)^2}       | r = \frac \d {\d x} \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) x^k       | c =  }} {{eqn | ll= \leadsto       | l = -\frac 2 {\left({1 + x}\right)^3}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) k x^{k - 1}       | c = differentiating {{WRT|Differentiation}} $x$ }} {{eqn | ll= \leadsto       | l = \frac 2 {\left({1 + x}\right)^3}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^{k - 1} \left({k + 1}\right) k x^{k - 1}       | c = taking one of the $-1$s out }} {{eqn | r = \sum_{k \mathop = 1}^\infty \left({-1}\right)^{k - 1} \left({k + 1}\right) k x^{k - 1}       | c = the term in $k = 0$ vanishes }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 2}\right) \left({k + 1}\right) x^k       | c = Translation of Index Variable of Product }} {{eqn | ll= \leadsto       | l = \frac 1 {\left({1 + x}\right)^3}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\left({k + 2}\right) \left({k + 1}\right)} 2 x^k       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\sqrt {1 + x} }       | r = \left({1 + x}\right)^{-\frac 1 2}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({-\frac 1 2}\right)^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({\left({-\frac 1 2}\right) - j}\right)} {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\displaystyle \prod_{j \mathop = 1}^k \left({2 j - 1}\right)} {2^k k!} x^k       | c = Translation of Index Variable of Product and simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\displaystyle \prod_{j \mathop = 1}^k \left({2 j - 1}\right) \prod_{j \mathop = 1}^k \left({2 j}\right)} {2^k k! \displaystyle \prod_{j \mathop = 1}^k \left({2 j}\right)} x^k       | c = multiplying top and bottom by $\displaystyle \prod_{j \mathop = 1}^k \left({2 j}\right)$ }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\displaystyle \prod_{j \mathop = 1}^{2 k} j} {2^k k! \left({2^k \displaystyle \prod_{j \mathop = 1}^k j}\right)} x^k       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\left({2 k}\right)!} {2^k k! \left({2^k k!}\right)} x^k       | c = {{Defof|Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\left({2 k}\right)!} {\left({2^k k!}\right)^2} x^k       | c = {{Defof|Factorial}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\sqrt {1 + x} }       | r = \left({1 + x}\right)^{\frac 1 2}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({\frac 1 2}\right)^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({\frac 1 2 - j}\right)} {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({1 - 2 j}\right)} {2^k k!} x^k       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({2 j - 1}\right)} {2^k k!} x^k       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^{k - 1} \frac {\displaystyle \prod_{j \mathop = 1}^{k - 1} \left({2 j - 1}\right)} {2^k k!} x^k       | c = extracting the case where $k = 0$ }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^{k - 1} \frac {\displaystyle \prod_{j \mathop = 1}^{k - 1} \left({2 j - 1}\right) \prod_{j \mathop = 1}^{k - 1} \left({2 j}\right)} {2^k k! \displaystyle \prod_{j \mathop = 1}^{k - 1} \left({2 j}\right)} x^k       | c = multiplying top and bottom by $\displaystyle \prod_{j \mathop = 1}^{k - 1} \left({2 j}\right)$ }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^{k - 1} \frac {\displaystyle \prod_{j \mathop = 1}^{2 \left({k - 1}\right)} j} {2^k k! \left({2^{k - 1} \displaystyle \prod_{j \mathop = 1}^{k - 1} j}\right)} x^k       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^{k - 1} \frac {\left({2 \left({k - 1}\right)}\right)!} {2^{2 k - 1} k! \left({k - 1}\right)!} x^k       | c = {{Defof|Factorial}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\sqrt [3] {1 + x} }       | r = \left({1 + x}\right)^{-\frac 1 3}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\left({-\frac 1 3}\right)^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\displaystyle \prod_{j \mathop = 0}^{k - 1} \left({\left({-\frac 1 3}\right) - j}\right)} {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \frac {\displaystyle \prod_{j \mathop = 1}^k \left({3 j - 1}\right)} {3^k k!} x^k       | c = Translation of Index Variable of Product and simplifying }} {{eqn | r = 1 - \frac 1 3 x + \frac {1 \times 4} {3 \times 6} x^2 - \frac {1 \times 4 \times 7} {3 \times 6 \times 9} x^3 + \cdots       | c = extracting the juice }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\sqrt [3] {1 + x} }       | r = \paren {1 + x}^{\frac 1 3}       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\paren {\frac 1 3}^{\underline k} } {k!} x^k       | c = General Binomial Theorem }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\frac 1 3 - j} } {k!} x^k       | c = {{Defof|Falling Factorial}} }} {{eqn | r = \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {1 - 3 j} } {3^k k!} x^k       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {3 j - 1} } {3^k k!} x^k       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^{k - 1} \frac {\ds \prod_{j \mathop = 1}^{k - 1} \paren {3 j - 1} } {3^k k!} x^k       | c = extracting the case where $k = 0$ }} {{eqn | r = \dfrac 1 {\sqrt [3] {1 + x} } = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots       | c = extracting the juice }} {{end-eqn}} {{qed}}
By definition of a power to a real number: :$a^x = \exp \left({x \ln a}\right)$ As $x \ln a$ is itself a real number, we can use Power Series Expansion for Exponential Function: {{begin-eqn}} {{eqn | ll= \forall x \in \R:       | l = \exp x       | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}       | c =  }} {{eqn | r = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots       | c =  }} {{end-eqn}} substituting $x \ln a$ for $x$. {{qed}}