Eden Solomon

2022-06-26

Limit of a harmonic subseries minus "its" logarithm
$\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\frac{1}{3k-1}-\frac{1}{3}\mathrm{ln}\left(n\right)$
I think that inserting the other terms and then subtracting them would not help. I need just the ideea. Thank you.

Quinn Everett

Expert

It is not very difficult to show that, for natural values of a, $\sum _{k=1}^{n}\frac{1}{n+a}={H}_{n+a}-{H}_{a}$ , where ${H}_{n}$ is the ${n}^{th}$ harmonic number. But in our case, $a=-\frac{1}{3}$, so we have to extend the formula or definition of ${H}_{n}$ so as to encompass non-natural arguments as well. Luckily for us, Euler has already done it some $250$ years ago: ${H}_{n}={\int }_{0}^{1}\frac{1-{x}^{n}}{1-x}dx$ . Using this, we can now compute the value of ${H}_{-\frac{1}{3}}$ to be $\frac{1}{2}\left(\frac{\pi }{\sqrt{3}}-3\mathrm{ln}3\right)$. Now use the fact that $\underset{n\to \mathrm{\infty }}{lim}{H}_{n}-\mathrm{ln}n=\gamma$. See here for more details on how to handle the integral in question.

pokoljitef2

Expert

If it were $\sum _{k=1}^{n}\frac{1}{k}-\mathrm{ln}n$ , we'd end up with $\gamma$. So I'd suggest to investigate
$\sum _{k=1}^{\mathrm{\infty }}\left(\frac{3}{3k-1}-\frac{1}{k}\right)$

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