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It is very well known that:
$$\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k}-\ln n = \gamma$$ Is there anything known about $$\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{2k}-\frac{\ln n}2 $$ or $$\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{2k-1}-\frac{\ln n}2 $$

Edit : Originally the question was : $\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{2k-1}-\ln n $ which of course is not interesting at all.

Did
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jimjim
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    Both the sums will be approximately $0.5 \ln n$, and so when you subtract $\ln n$ and take the limit, you get something that diverges to $-\infty$. – Aaron Sep 25 '15 at 06:43
  • @Aaron : good point, revising the question to reflect a deeper question. – jimjim Sep 25 '15 at 06:56
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    For the revised problems, the first one you can simply factor out $1/2$ to get $\gamma/2$. I'm not sure off the top of my head how to tackle the second one, though. – Aaron Sep 25 '15 at 07:03
  • @Aaron : again, 100% correct, I should have seen that. But I leave it as it is. I have seen a relationship between sum of odd inverse squares and sum of even inverse squares. – jimjim Sep 25 '15 at 07:06
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    Fair point. With that insight, I think I have a solution. – Aaron Sep 25 '15 at 07:11
  • A related question. – Lucian Sep 25 '15 at 10:57

2 Answers2

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Let $A_n = \sum\limits_{k=1}^{n}1/(2k), B_n = \sum\limits_{k=1}^{n}1/(2k-1).$ Then $A_n + B_n -\ln (2n) \to \gamma.$ This gives

$$A_n + B_n - (\ln 2 +\ln n) \to \gamma,$$ hence $$[A_n - (\ln n)/2] + [B_n - (\ln n)/2 -\ln 2] \to \gamma.$$

We know that $[A_n - (\ln n)/2] \to \gamma/2,$ so it follows that $[B_n - (\ln n)/2 -\ln 2] \to \gamma/2.$ Thus $$B_n - (\ln n)/2 \to \gamma/2 + \ln 2.$$

Did
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zhw.
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We wish to evaluate $\displaystyle\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{2k-1} - \frac{\ln n}{2}$. Let us take a a given that the limit exists. Since $$\gamma=\displaystyle\lim_{n\to\infty}\sum_{k=1}^{2n} \frac{1}{2k} -\ln 2n=\lim_{n\to\infty}\sum_{k=1}^{n} \frac{1}{2k} +\sum_{k=1}^{n} \frac{1}{2k-1} -\ln n - \ln 2,$$

we can subtract off the sum of even terms to get $\gamma/2 + \ln 2$.

Aaron
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