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Evaluate $\displaystyle\lim_{x\to 0+}\sum_{n=2020}^{\infty}\frac{(-1)^n}{\log^x(n)}$.

It's easy to check that $$ \lim_{x\to 0+}\log^x(n)=1. $$

And for fixed $x$, since this is a decreasing alternating series, it converges obviously.

Then I don't know how to continue.

Knt
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    "since this is a decreasing alternating series, it converges absolutely." Erm, no. You mean conditionally. – Clement C. Aug 03 '19 at 16:37
  • Are you looking for a closed-from expression for this? Because I would be surprised if that exists. – QC_QAOA Aug 03 '19 at 16:39
  • @ClementC. It's a typo. I meant obviously. – Knt Aug 03 '19 at 17:02
  • @NickGuerrero I have tried so. But I couldn't find it. – Knt Aug 03 '19 at 17:03
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    So, if I may, why to you believe it has a closed-form? Where is the statement coming from? And you're expecting a closed-form (as a function of $x$) for every fixed $x>0$, not limit when $x$ goes to say $\infty$? – Clement C. Aug 03 '19 at 17:03
  • The only way I can really make sense of the question (unless you have a magical guarantee that the closed-form is nice, for every fixed $x>0$, which I strongly doubt) is that you are actually asked to compute either $\lim_{x\to \infty} \sum_{n=2020}^\infty [\dots]$, or the asymptotics as $xto \infty$. (Or as $x\to 0^+$, but I would find that less likely.) – Clement C. Aug 03 '19 at 17:13
  • This question comes from internet in my country, which claims it's a problem in a practice graduate entrance test. And it's based on student's memory. So I also doubt if it's reasonable. – Knt Aug 03 '19 at 17:17
  • Yes, I am 99% sure you are asked the limit of the sum when $x\to a$, for $a$ either $0^+$ or $\infty$. Which one, I don't know, but as stated this is very unlikely to be the correct question. – Clement C. Aug 03 '19 at 17:20
  • Now I have corrected. Can you give an answer? – Knt Aug 03 '19 at 17:23

1 Answers1

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Let $f(t) = (\log t)^{-x}$ then $$\sum_{n \ge 2N} (-1)^n f(n) =-\sum_{n \ge N} (f(2n+1)-f(2n)) = - \sum_{n \ge N} \int_0^1 f'(2n+t)dt \\= - \sum_{n \ge N} \int_0^1 (f'(2n+2t) + O(|f''(2n)|))dt = \frac12 f(2N) + O( \sum_{n \ge N} |f''(2n)|)\\ = \frac12 (\log 2N)^{-x} +O(\frac{x (\log N)^{-x-1}}{N})$$ as $f'(t) = \frac{-x (\log t)^{-x-1}}{t}, f''(2n) = O(\frac{x (\log n)^{-x-1}}{n^2})$

reuns
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  • $n\leadsto N$ in the last line. The asymptotics are with regard to $N$? But $N$ is fixed in the question, it's $N=1010$. If there are asymptotics to take, that'd be with regard to $x$. – Clement C. Aug 03 '19 at 17:10
  • That's not what I mean. When $x\to \infty$ for $N$ constant, your asymptotics do not make sense, as $x C^{x-1}$ (which is in the $O(\cdot)$ as low-order) dominates $C^x$ (which you write as dominating term). You took asymptotics for $x$ fixed and $N\to \infty$. – Clement C. Aug 03 '19 at 17:15
  • Do you even read what I write? $|x| \leq 1/4$ is unlikely to hold if the focus is the asymptotic behavior when $x\to\infty$. – Clement C. Aug 03 '19 at 17:16
  • At this point, i strongly believe the question is not well-posed, incidentally (see my comment below it), and that the OP want the answer as either $x\to\infty$ or $x\to 0^+$. However, your answer addresses neither, as you consider the asymptotic behavior as $N\to\infty$ (while in the question, $2N=2020$ is a constant). – Clement C. Aug 03 '19 at 17:18
  • No I don't consider the asymptotic as $N \to \infty$, I consider the obvious estimate of the tail, with uniform O-constant. – reuns Aug 03 '19 at 17:21
  • I'm not going to continue here, but note (1) The OP has changed the question (may change it again?), and (2) you have edited your answer since, making some of my comments above (specifically, the one about $C^x$ v. $xC^{x-1}$) no longer accurate. – Clement C. Aug 03 '19 at 17:25
  • Yes I changed the sign of $x$ to make it similar to OP and to make it clear that in your comment there is no reason to consider its behavior as $x \to -\infty$. At first I didn't care of the sign because I was thinking to $x \to 0$ – reuns Aug 03 '19 at 17:27