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This question already has an answer here:

I would like to know how to find out the sum of this series:

$$1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \frac{1}{5^2} - \frac{1}{6^2} + \cdots$$

The answer is that it converges to a sum between $\frac 34$ and $1$, but how should we go about estimating this sum?

Thanks!

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marked as duplicate by user147263, apnorton, colormegone, hardmath, Claude Leibovici Aug 28 '14 at 5:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ math.stackexchange.com/questions/430973 $\endgroup$ – David Aug 28 '14 at 0:10
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    $\begingroup$ Which do you want, to estimate the sum or find the exact value? $\endgroup$ – Robert Israel Aug 28 '14 at 0:16
  • $\begingroup$ An explicit formula for your sum can be obtained quickly from Euler's famous result $1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6}$. That result has many proofs, none really simple. Nowadays it is usually derived as a consequence of certain facts about Fourier series. $\endgroup$ – André Nicolas Aug 28 '14 at 0:29
  • $\begingroup$ See Basel problem. $\endgroup$ – Lucian Aug 28 '14 at 0:31
  • $\begingroup$ Note: this is $\eta(2)$, where $\eta$ is the Dirichlet eta function. Just in case anyone wanted to ask about another value. $\endgroup$ – apnorton Aug 28 '14 at 2:10
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The sum of this series is $\frac{\pi^2}{12}$.

Explantion

We already know that:

$$1+ \frac{1}{2^2} + \frac{1}{3^2} + \text{...} = \frac{\pi^2}{6}$$

HINT

Note that:

$$\large \frac{-1}{2^2} = \frac{1}{2^2} - \frac{1}{2^2} - \frac{1}{2^2}$$

Now,

$$1- \frac{1}{2^2} + \frac{1}{3^2} - \text{...} = \left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right) - 2 * \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + ...\right)$$

$$ = \frac{\pi^2}{6} - \frac{2}{2^2}\left(\frac{1}{1} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ...\right)$$

$$ = \frac{\pi^2}{6} - \frac{1}{2}*\frac{\pi^2}{6}$$

$$= \frac{\pi^2}{12}$$

Comment if you have questions.

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  • $\begingroup$ How do you know that? $\endgroup$ – inggumnator Aug 28 '14 at 0:08
  • $\begingroup$ @inggumnator I'll post shortly the solution. $\endgroup$ – Varun Iyer Aug 28 '14 at 0:08
  • $\begingroup$ @inggumnator do you understand now how it is derived? I hope this helps. $\endgroup$ – Varun Iyer Aug 28 '14 at 0:15
  • $\begingroup$ @VarunIyer: Nice derivation. It might be worthwhile to mention that the rearrangement of the series is valid because it is absolutely convergent (to $\pi^2/6$, as you noted). $\endgroup$ – Bungo Aug 28 '14 at 0:21
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    $\begingroup$ @inggumnator yes. If you notice we already have a multiple of $2$ outside the parenthesis. The common term here is $\frac{1}{2^2}$, so we pull that out as well to get our sum $\frac{\pi^2}{6}$ again. Thus the number outside the parenthesis now is $\frac{2}{2^2}$. $\endgroup$ – Varun Iyer Aug 28 '14 at 1:24
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It is well known that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.

Thus, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \sum_{m=1}^{\infty}\dfrac{1}{(2m)^2} = \dfrac{1}{4}\sum_{m=1}^{\infty}\dfrac{1}{m^2} = \dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{24}$.

Hence, $\displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} = \sum_{n=1}^{\infty}\dfrac{1}{n^2} - \sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{8}$.

Finally, $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^2} = \displaystyle\sum_{\substack{n=1\\ n \ \text{is odd}}}^{\infty}\dfrac{1}{n^2} - \displaystyle\sum_{\substack{n=1\\ n \ \text{is even}}}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{8} - \dfrac{\pi^2}{24} = \dfrac{\pi^2}{12}$.

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If we did not know an exact answer, here is a crude way to estimate the sum of the series.

Our series is an alternating series, with terms that decrease in absolute value and have limit $0$. So the error made by truncating at a particular term has absolute value less than the absolute value of the first "neglected" term.

For example, if we use $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\frac{1}{5^2}$ as an approximation, then the absolute value of the error is less than $\frac{1}{6^2}$. Furthermore, the error is "negative," that is, our estimate is greater than the true value.

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  • $\begingroup$ Thanks, this is a nifty trick! $\endgroup$ – inggumnator Aug 28 '14 at 0:53
  • $\begingroup$ You are welcome. For the kind of series described (signs alternate, terms go down in absolute value, approach $0$) the truncation error is easy to estimate. $\endgroup$ – André Nicolas Aug 28 '14 at 3:33
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HINT:

This can be written as $$\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i^2}$$

$$$$

Setting $\displaystyle a_i= \frac{(-1)^{i+1}}{i^2}$:

$$\sum_{i=1}^{\infty} a_{2i-1}=\sum_{i=1}^{\infty} \frac{1}{(2i-1)^2}$$

$$\sum_{i=1}^{\infty} a_{2i}=-\frac{1}{4}\sum_{i=1}^{\infty} \frac{1}{i^2}$$

When the series $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}$ converge then the series $\displaystyle \sum_{i=1}^{\infty} a_{i}$ also converges.

$$\sum_{i=1}^{\infty} a_{i}=\sum_{i=1}^{\infty} a_{2i-1}+\sum_{i=1}^{\infty} a_{2i}$$

Find where the series $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}$ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}$ converge.

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    $\begingroup$ Sorry why does finding where the two series converge show where the original series converge? $\endgroup$ – inggumnator Aug 28 '14 at 0:40
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    $\begingroup$ When $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}=x $ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}=y$ then $$\sum_{i=1}^{\infty} a_{i}=x+y$$ $\endgroup$ – user159870 Aug 28 '14 at 0:43

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