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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!

Community
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thbcm
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4 Answers4

11

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.

Varun Iyer
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  • How do you know that? – thbcm Aug 28 '14 at 00:08
  • @inggumnator I'll post shortly the solution. – Varun Iyer Aug 28 '14 at 00:08
  • @inggumnator do you understand now how it is derived? I hope this helps. – Varun Iyer Aug 28 '14 at 00:15
  • @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). –  Aug 28 '14 at 00:21
  • @AndréNicolas yes it can, as I have showed here. – Varun Iyer Aug 28 '14 at 00:28
  • @VarunIyer: I placed the comment in the wrong place, and have shifted it tp the right place. The comment is about the Euler result $\sum \frac{1}{n^2}=\frac{\pi^2}{6}$, which is not proved in any of the answers. – André Nicolas Aug 28 '14 at 00:32
  • Why did you multiple $(\frac 11 + \frac {1}{2^2} + \frac {1}{3^2})$ by $\frac {2}{2^2}$? Isn't that like dividing by 2, when it should be multiplying by 2? – thbcm Aug 28 '14 at 00:33
  • Oh wait is it because the expression in the first line has half the terms of the second, since it's missing the fractions with odd-numbered denominators? – thbcm Aug 28 '14 at 00:35
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    @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 $\frac{2}{2^2}$. – Varun Iyer Aug 28 '14 at 01:24
5

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

JimmyK4542
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3

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.

André Nicolas
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0

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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    Sorry why does finding where the two series converge show where the original series converge? – thbcm Aug 28 '14 at 00:40
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    When $\displaystyle \sum_{i=1}^{\infty} a_{2i-1}=x $ and $\displaystyle \sum_{i=1}^{\infty} a_{2i}=y$ then $$\sum_{i=1}^{\infty} a_{i}=x+y$$ –  Aug 28 '14 at 00:43