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I would like to analyze the convergence of the expression

$$a_n =\frac {\sum_{k=1}^n \frac{1}{k} - (1-\frac{2}{n})\ln n}{\ln n}$$

as $n \to \infty$. In particular, I'd like to know that, if the limit exists and is finite, whether it is $0$. The terms in the numerator look similar to the limit of $\sum_{k=1}^n \frac{1}{k} - \ln n$, which determines the Euler-Mascheroni constant. However, that extra factor of $(1 - \frac{2}{n})$ in front of $\ln n$ is making the numerator just a little bit larger than $\sum_{k=1}^n \frac{1}{k} - \ln n$.

So, the question is, even though the numerator might go to infinity because of the $(1 - \frac{2}{n})$ factor, does the $\ln n$ in the denominator still have enough force to send the whole expression to $0$? So far, I have only tried an "experimental" method, making a large spreadsheet and computing values up to $n = 40$. I am convinced that the $a_n$ are decreasing monotonically, but even by computing so many values I can't make out whether the limit will be positive or $0$. Guess I'll have to turn to analysis for some help.

Hints or solutions are greatly appreciated.

cskhr
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JZS
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1 Answers1

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Why don't you instead approximate: $$\sum_{k=1}^n \frac{1}{k} \simeq \gamma+\ln(n) $$ After that expansion, and some algebra will show that limit is 0. i.e. you get $$a(n)=\frac{\gamma}{\ln(n)}+\frac{2}{n}\rightarrow 0$$