$$\sum^{\infty}_{n=1} \frac{(-1)^n}{n^2}$$

I know how to prove this one is convergent by alternating test. like Let $b_n = \frac{1}{n^2}$ and limit of this sum is $0$ and also decreasing. However, my professor doesn't allow me to use that I did not learn yet. So I cannot use alternating test(I can use p-series, comparison, ratio test). I don't know how to solve this one..... without alternating test.

  • $\begingroup$ Consider the largest possible value this could be, i.e., do a bounding analysis. For example, what if you just look at a bound on the magnitude? $\endgroup$ – Michael Mar 31 '17 at 1:19
  • $\begingroup$ Could compare to $\frac{(-1)^n}{n}$ (although you probably used alternating series test for that one too) $\endgroup$ – user12345 Mar 31 '17 at 1:21
  • $\begingroup$ Which tests have you learned so far? $\endgroup$ – mephistolotl Mar 31 '17 at 1:22
  • 3
    $\begingroup$ You can use the fact that an absolutely convergent series is conditionally convergent and just use the p-test $\endgroup$ – B.A Mar 31 '17 at 1:22
  • $\begingroup$ Could you show me how to use the fact that "an absolutely convergent series is conditionally convergent" $\endgroup$ – Kwangi Yu Mar 31 '17 at 1:31

You know about $p$-series, so you know the series $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.

Proving convergence of your alternating series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ is equivalent to proving the sequence of its partial sums, $(S_n)_{n\geq 1}$ is Cauchy.

It will be helpful to compare it to the $p$-series you do know something about; so observe that for any $m\geq 1$ ,$p\geq 0$, $$ \left\lvert\sum_{n=m+1}^{m+p} \frac{(-1)^n}{n^2} \right\rvert \leq \sum_{n=m+1}^{m+p} \left\lvert\frac{(-1)^n}{n^2} \right\rvert = \sum_{n=m+1}^{m+p} \frac{1}{n^2} \tag{1} $$ by the triangle inequality.

So fix $\varepsilon > 0$. Since $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, it is Cauchy, so there exists $m\geq 1$ such that, for all $p\geq 0$, $\sum_{n=m+1}^{m+p} \frac{1}{n^2} \leq \varepsilon$. By the above (1), for the same $m$ and any $p\geq 0$, we have $$ \lvert S_{m+p}-S_m\rvert = \left\lvert\sum_{n=m+1}^{m+p} \frac{(-1)^n}{n^2} \right\rvert \leq \varepsilon. $$

This shows $(S_n)_{n\geq 1}$ is Cauchy, thus convergent.

  • $\begingroup$ oh... I got it Thank you so much !!!!!!! $\endgroup$ – Kwangi Yu Mar 31 '17 at 1:34
  • $\begingroup$ @KwangiYu You're welcome! Basically, this is a special case of the argument shoing "absolute convergence implies conditional convergence" -- the trick is to use Cauchyness. $\endgroup$ – Clement C. Mar 31 '17 at 1:35
  • 2
    $\begingroup$ Assume $\sum_{n}^\infty |a_n|$ converges. Note that we have $0\le a_n+|a_n|\le 2|a_n|$. Then, we can write $$\sum_{n}^N a_n=\sum_{n}^N (a_n+|a_n|)-\sum_{n}^N |a_n|$$which shows that the series of interest is the sum of two convergent series. Done. No need to bring in Cauchy. $\endgroup$ – Mark Viola Mar 31 '17 at 4:30
  • $\begingroup$ @Dr.MV Good point. A "proof from the book" (yet, probably not the most intuitive to a student new to the subject...) $\endgroup$ – Clement C. Mar 31 '17 at 12:35
  • $\begingroup$ CC, I'm not sure which is the less intuitive. I beleve that appealing to Cauchy sequences seem a bit more advanced and less intuitive than appealing to straightforward dominance arguments. $\endgroup$ – Mark Viola Mar 31 '17 at 13:33

Note that we can write

$$\begin{align} \sum_{n=1}^{2N} \frac{(-1)^n}{n^2}&=\sum_{n=1}^{2N} \frac{1+(-1)^n}{n^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\\\\ &=\sum_{n=1}^N\frac{2}{(2n)^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\\\\ &=\frac12 \sum_{n=1}^N\frac{1}{n^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\tag1 \end{align}$$

Since we know that both of the partial sums on the right-hand side of $(1)$ converge, then the partial sum on the left-hand side must converge also. In fact, we have from $(1)$

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=-\frac12 \sum_{n=1}^\infty \frac{1}{n^2}$$

And that completes the proof.

We have a much more general way to approach this problem. We will now show that if a series converges absolutely, then it converges.

To show this, we assume that $\sum_{n=1}^\infty |a_n|$ converges. Now, since $\sum_{n=1}^\infty |a_n|$ converges, then $\sum_{n=1}^\infty 2|a_n|$ converges also.

Note that we have the inequalities $0\le a_n+|a_n|\le 2|a_n|$. Since the sequence $a_n+|a_n|$ is non-negative, then

$$0\le \sum_{n=1}^N(a_n+|a_n|)\le \sum_{n=1}^N2|a_n| \tag 2$$

The sequence of partial sums $S_N=\sum_{n=1}^N(a_n+|a_n|)$ in $(2)$ is monotonically increasing and bounded above by $\sum_{n=1}^\infty 2|a_n|$. Therefore, $S_N$ converges.

Finally, we can write

$$\sum_{n=1}^N a_n=\sum_{n=1}^N (a_n+|a_n|)-\sum_{n=1}^N |a_n| \tag 3$$

Inasmuch as the sequence of partial sums on the right-hand side of $(3)$ converge, the sequence of partial sums on the left-hand side of $(3)$ converge also.

We have just shown that any absolutely convergent sequence also converges.

  • $\begingroup$ Kwangi, please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Mar 31 '17 at 23:07

$$\sum \frac{(-1)^n}{n^2} = -1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} \cdots = \sum \frac{1}{(2n)^2}- \frac{1}{(2n-1)^2} = \frac{1}{4}\sum\frac{1}{n^2} - \sum \frac{1}{(2n-1)^2}$$

The latter two sums are convergent by $p$ series test.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.