=== Necessary Condition === We will first show that: :if the condensed series $\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges then: :$\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges as well. Assume $\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges. Consider the graph of $\sequence {a_n}$ and the partial sums of $\sum 2^n \map a {2^n}$: :700px The dotted black line represents the sequence $\sequence {a_n}$. The $n$th rectangle has: :Base equal to $2^n$ :Altitude equal to $\map a {2^n}$ The series of partial sums of the areas of the rectangles are: :$\displaystyle \sum_{n \mathop = 1}^N 2^n \map a {2^n}$ which is precisely the defined condensed series. The series $\sum a_n$ can be viewed as the sum of the areas of rectangles with width $1$ and height $a_n$.  The diagram suggests that the partial sums of $\sum a_n$ are not greater than the condensed partial sums: :$\displaystyle \sum_{n \mathop = 1}^N a_n \le \sum_{n \mathop = 1}^N 2^n \map a {2^n}$ To formalize this claim, observe that because $\sequence {a_n}$ is decreasing: :$a_1 + \underbrace {a_2 + a_3}_{\le 2 a_2} + \underbrace {a_4 + a_5 + a_6 + a_7}_{\le 4 a_4} + \cdots + a_N \le a_1 + 2 a_2 + 4 a_4 + \cdots + 2^N \map a {2^N}$ As $n \to +\infty$, the {{RHS}} converges by hypothesis. Hence $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ also converges, by the Comparison Test. {{qed|lemma}} === Sufficient Condition === We will show that if $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges, the condensed series $\displaystyle \sum_{n \mathop = 1}^\infty 2^n \map a {2^n}$ converges as well. Assume $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges. From the Combination Theorem for Sequences, $2\displaystyle \sum_{n \mathop = 1}^\infty a_n = \displaystyle \sum_{n \mathop = 1}^\infty 2 a_n$ converges as well. Consider, then, the graph of $\sequence {a_n}$ and of the partial sums of what will be shown to equal $\displaystyle \frac 1 2 \sum 2^n \map a {2^n}$: :700px The dotted black line represents the sequence $\sequence {a_n}$. The $n$th rectangle has: :Base equal to $2^n$ :Altitude equal to $\map a {2 \cdot 2^n}$ The series of partial sums of the areas of the rectangles are: {{begin-eqn}} {{eqn | o =       | r = \sum_{n \mathop = 1}^N 2^n \map a {2 \cdot 2^n} }} {{eqn | r = \sum_{n \mathop = 1}^N 2^n \map a {2^{n + 1} } }} {{eqn | r = \sum_{n - 1 \mathop = 1}^N 2^{n - 1} \map a {2^n}       | c = replacing $n$ with $n - 1$ }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 2}^N 2^n \map a {2^n} }} {{end-eqn}} From Deletion of Terms from a Sequence, it is justified to examine only the behavior of $n$ for $n \ge 2$. The series $\sum a_n$ can be viewed as the sum of the area of rectangles with height $a_n$ and width $1$. The diagram suggests that the partial sums of $\sum a_n$ satisfy the following inequality: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 2}^N a_n       | o = \ge       | r = \frac 1 2 \sum_{n \mathop = 2}^N 2^n \map a {2^n}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \displaystyle \sum_{n \mathop = 2}^N 2 a_n       | o = \ge       | r = \sum_{n \mathop = 2}^N 2^n \map a {2^n}       | c =  }} {{end-eqn}} To formalize this claim, observe that because $\sequence {a_n}$ is decreasing:  {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 2}^N 2a_n       | r = a_2 + \underbrace {a_2 + a_3}_{\ge 2 a_4} + \underbrace {a_3 + 2 a_4 + a_5}_{\ge 4 a_8} + \underbrace {a_5 + 2 a_6 + 2 a_7 + 2 a_8 + a_9}_{\ge 8 a_{16} } + \cdots + 2 a_N }} {{eqn | o = \ge        | r = a_2 + 2 a_4 + 4 a_8 + 8 a_{16} + \cdots + 2^n \map a {2^n} }} {{end-eqn}} By hypothesis, the {{LHS}} converges as $n \to +\infty$. Hence the condensed series converges as well, by the Comparison Test. {{qed}} {{Namedfor|Augustin Louis Cauchy|cat = Cauchy}}