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Is there someone who can show me how do I evaluate this sum :$$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$$

Note : In wolfram alpha show this result and in the same time by ratio test it's not a convince way to judg that is convergent series

Thank you for any help

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  • $\begingroup$ As a general remark, I think it should be mentioned that the Ratio Test is often "useless" for series with general terms which are ratios of polynomials. You can't even use it to tell if $ \ \sum_{n=1}^{\infty} \ \frac{1}{n^2} \ $ converges... $\endgroup$ – colormegone Jul 27 '15 at 3:14
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The parity of $\frac{n(n-1)}{2}$ is 4-periodic. Thus the sequence $(-1)^{\frac{n(n-1)}{2}}$ equals to: $$ 1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1, \, 1, \, -1, \, -1, \, 1 , \cdots$$ The original series' partial sum truncated at $N$ equals to $$ \sum_{k=0}^{K} \left( \frac{1}{4k+1} - \frac{1}{4k+2} - \frac{1}{4k+3} + \frac{1}{4k+4}\right) + \sum_{i=1}^{N - 4K - 4}\frac{(-1)^{\frac{i(i-1)}{2}}}{4K + 4 + i}$$ where $K = \lfloor \frac{N}{4}\rfloor - 1$.

Then by a discussion on the partial sum we can conclude that the series is convergent.

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  • $\begingroup$ Looking to the original series, you would get $-1, -1, +1, +1$... $\endgroup$ – johannesvalks Jul 27 '15 at 1:56
  • $\begingroup$ My mistake. Vote up! $\endgroup$ – johannesvalks Jul 27 '15 at 2:17
  • $\begingroup$ Why is it ok to group the terms like this? Take the series 1 -1 +1 -1 ... for example. If you grouped in pairs you'd get $(1-1)+(1-1) ...= 0 + 0 + 0+...$ so you'd conclude that the series converged to $0$ which is untrue. $\endgroup$ – lulu Jul 27 '15 at 2:30
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    $\begingroup$ @lulu Here the terms being grouped individually tend to $0$. This whole argument shows that there is a subsequence of the partial sums that converges. Any term of the general sequence of partial sums is within three terms of this convergent subsequence, and those three terms get arbitrarily small. The same kind of logic goes into the altenating series test. $\endgroup$ – alex.jordan Jul 27 '15 at 2:45
  • $\begingroup$ @alex.jordan Ok, thanks. I just worked out (and posted) the "Leibniz test" version of a solution, but I can see where this explicit argument gives you a hold on the remainder. Good! $\endgroup$ – lulu Jul 27 '15 at 2:50
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Some of the posted answers are using rearrangements or regrouping, but these are treacherous techniques to apply to series which diverge absolutely (as this one plainly does).

Leibniz' Theorem as usually stated only applies to alternating series, but we can modify the proof of Leibniz' Theorem to cover this particular series.

Leibniz Theorem: A series of the form $\sum(-1)^n a_n$ wherein the $a_n$ are all positive and the $a_n$ decrease to 0 converges.

Proof: Let $S_k$ denote the sum of the first k terms. Now look at two consecutive partial sums, $S_k$ and $S_{k+1}$ The next one, $S_{k+2}$, clearly falls between these (as the series is alternating and decreasing). In this way we see that all subsequent partial sums fall between any two consecutive partial sums. As the gap between consecutive partial sums goes to 0 we see that the partial sums, and therefore the series, must converge.

As this series is not alternating, we can't invoke the theorem casually. But the logic of the proof works just fine: for the given series we just observe that all subsequent partial sums are trapped between partial sums of the form $S_{4k+1}$ and $S_{4k+3}$ and the gaps between partial sums a fixed distance apart clearly goes to $0$.

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  • $\begingroup$ There is something sort of "Cauchy" about this argument, no? Thank you for taking me to task -- I was being a bit cavalier about "derangement". $\endgroup$ – colormegone Jul 27 '15 at 3:27
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    $\begingroup$ Personally, I never trust conditionally convergent series. There is something dishonest about them. If a series can't converge absolutely then I prefer it to diverge. I understand that my preferences may not signify much, but still. $\endgroup$ – lulu Jul 27 '15 at 3:31
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After $ \ n = 1 \ $ , the exponents of (-1) are binomial coefficients which are "double-alternating" between even and odd integers. So the series looks like

$$ 1 \ - \frac{1}{2} \ - \ \frac{1}{3} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ - \ \frac{1}{6} \ - \ \frac{1}{7} \ + \ \ldots \ \ . $$

As lulu I think properly objected to my separation of terms originally, since the absolute series here is the harmonic series, the "Riemann derangement" theorem may have something to say against it. It is perhaps safer then to group the terms as

$$ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left( \ \frac{1}{2k} \ + \ \frac{1}{2k+1} \ \right) \ = \ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left[ \ \frac{4k + 1}{2k \ (2k+1)} \ \right] $$ $$ = \ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left[ \ \frac{4k \ + \ 1}{4k^2 \ + \ 2k} \ \right] \ \ . $$

The term $ \ b_k \ = \ \frac{4k \ + \ 1}{4k^2 \ + \ 2k} \ $ proves to be monotonically decreasing toward zero for $ \ k \ \ge \ 1 \ $ , so the series converges by the alternating-series test. (And now my post looks a lot more like corindo's ...)

EDIT: Having thought about this a bit more, it occurred to me that I could have gathered this up a bit more (and now it comes still closer to what corindo is describing). The alternating series could have simply been written as

$$ 1 \ + \ \sum_{k=1}^{\infty} \ \left( \ \frac{-1}{4k-2} \ + \ \frac{-1}{4k-1} \ + \ \frac{1}{4k} \ + \ \frac{1}{4k+1} \ \right) $$ $$ = \ 1 \ - \ \sum_{k=1}^{\infty} \ \left[ \ \frac{32k^2 \ - \ 8k \ - \ 1}{4 \ x \ (2x-1) \ (4x-1) \ (4k+1)} \ \right] $$ $$ = \ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \left( \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ \right) \ \ $$ (with a little help from WolframAlpha). This removes the "alternating-sign" behavior and allow us to apply the "limit comparison test" against $ \ \sum_{k=1}^{\infty} \ \frac{1}{k^2} \ $ .

We can use this to set bounds on the sum (I've only pursued this a little at this point). It is not hard to show that $ \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ < \ \frac{2}{k^2} \ $ , from which we obtain

$$ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \left( \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ \right) \ > \ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \ \frac{2}{ k^2 } $$ $$ = \ 1 \ - \ \frac{1}{2} \ \zeta(2) \ = \ 1 \ - \ \frac{\pi^2}{12} \ \ . $$

Others here are likely better able to set tighter bounds on the sum than I am at the moment.

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  • $\begingroup$ As the series diverges absolutely, working with a rearrangement doesn't easily imply anything about the original. $\endgroup$ – lulu Jul 27 '15 at 2:26
  • $\begingroup$ @lulu Gaah, that's a quite reasonable objection. I'll see if I can find a better way to bound this series and delete this post in a little while... $\endgroup$ – colormegone Jul 27 '15 at 2:42
  • $\begingroup$ All right, I decided to "gut" the post instead; hopefully, what I have now works properly... $\endgroup$ – colormegone Jul 27 '15 at 3:16
  • $\begingroup$ And now I agree with the proof. And I agree that it's more or less equivalent to corindo's (and to mine, for that matter). But it seems to me that the question is hard enough to justify several variants of the argument. $\endgroup$ – lulu Jul 27 '15 at 3:20
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I post this answer because Dirichlet's test has not been mentioned in any of the previous answers. Let $a_n=(-1)^{n(n-1)/2}$ and $b_n=1/n$. The partial sums of $a_n$ are bounded and $b_n$ is decreasing and converging to $0$. Dirichlet's test implies the series is convergent.

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Let $A_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}.$

Let $B_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n}.$

Let $C_N$ be the $N$th partial sum of $\sum_{n=1}^{\infty}\frac{(-1)^{n(n-1)/2}}{n}.$

Then $C_{2N} = A_N + B_N.$ By the alternating series test, both $A_N,B_N$ converge. Hence $C_{2N}$ converges to some limit $S$. Now $C_{2N+1}=C_{2N+1}-C_{2N}+C_{2N}$ and $C_{2N+1}-C_{2N}=(\pm 1)/(2N+1) \to 0.$ Thus $C_{2N+1}\to S.$ A sequence whose even terms converge to $S$ and whose odd terms converge to $S$ itself converges to $S.$ Thus $C_N$ converges, which is the desired result.

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