Step by step solution to "Proving that:∑n=1∞(−1)n+1n2=π212"

sunyerneq

Answered question

2022-02-23

Proving that:

\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n^{2}} = \frac{π^{2}}{12}

Ian Adams

Skilled2022-03-07Added 163 answers

Split it:

\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n^{2}} = \sum_{n o d d} \frac{1}{n^{2}} - \sum_{n e v e n} \frac{1}{n^{2}}

Add and subtract the "even" part:

\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n^{2}} = (\sum_{n o d d} \frac{1}{n^{2}} + \sum_{n e v e n} \frac{1}{n^{2}}) - \sum_{n e v e n} \frac{1}{n^{2}} - \sum_{n e v e n} \frac{1}{n^{2}}

= \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - 2 \sum_{n e v e n} \frac{1}{n^{2}} = \frac{π^{2}}{6} - 2 \sum_{n e v e n} \frac{1}{n^{2}}

Now, notice that:

\sum_{n e v e n} \frac{1}{n^{2}} = \sum_{i = 1}^{\infty} \frac{1}{(2 i)^{2}} = \frac{1}{4} \sum_{i = 1}^{\infty} \frac{1}{i^{2}} = \frac{1}{4} \frac{π^{2}}{6}

Therefore:

\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n^{2}} = \frac{π^{2}}{6} - 2 \frac{1}{4} \frac{π^{2}}{6} = \frac{π^{2}}{12}

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