I want to determine the limit of the following sequence


From the foregoing, consider


Also try to consider Stirling's approximation, so you would have to find the limit of $$x_n=\frac{1}{n}\left(\sum_{k=1}^n\frac{n+1-k}{k}-\log(\sqrt{2πn}\left(\frac{n}{e}\right)^n\right)$$

I don't know if my previous statement is completely true, besides, from this expression it is difficult for me to find the requested limit.

Any help please?

  • 4
    $\begingroup$ $\sum_{k=1}^n\frac{n+1-k}{k}=(n+1)H_n-n$, where $H_n=\sum_{k=1}^n\frac{1}{k}$ - harmonic number. Using the asymptotics of $H_n=\ln n+\gamma+O(\frac{1}{n})$ you can find the desired limit en.wikipedia.org/wiki/Harmonic_number $\endgroup$
    – Svyatoslav
    Jan 14, 2022 at 14:21
  • $\begingroup$ You are missing an $o(1)$. Replacing the $\log(n!)$ with Stirling approximation you need to acknowledge the error: $\log(n!) = \log(\sqrt{2\pi n} (\frac{n}{e})^n) + o(1)$ $\endgroup$
    – jjagmath
    Jan 14, 2022 at 14:25
  • $\begingroup$ Use this with $a_n:=nx_n,\,b_n:=n$. $\endgroup$
    – J.G.
    Jan 14, 2022 at 14:37
  • $\begingroup$ Well, it seems that it would be convenient for me to evaluate the limit of $\frac{1}{n}((n+1)(\log(n)+\gamma)-n-(n\log(n) - n))$. However, I can't find an answer that reassures me, any help for the calculation of this last limit? $\endgroup$
    – Wrloord
    Jan 14, 2022 at 15:06

2 Answers 2


Let $x_n$ be the sequence given by

$$\begin{align} x_n&=\frac1n \left(\sum_{k=1}^n \frac{n+1-k}{k}-\log(k)\right)\\\\ &=\frac1n \left(\sum_{k=1}^n \frac nk+\frac1k-1-\log(k)\right)\\\\ &=-1+\underbrace{\frac1n\sum_{k=1}^n \frac1k}_{\to 0\,\,\text{as}\,\,n\to \infty} +\underbrace{\sum_{k=1}^n \frac1k-\log(n)}_{\to \gamma\,\,\text{as}\,\,n\to \infty}-\underbrace{\frac 1n\sum_{k=1}^n \log(k/n)}_{\to -1\,\,\text{as}\,\,n\to \infty}\tag1 \end{align}$$

Therefore, we find that

$$\lim_{n\to \infty}x_n=\gamma$$

  • 3
    $\begingroup$ The last term should be $-\dfrac{1}{n} \sum_{k=1}^{n} \log \left( \dfrac{k}{n} \right)$. $\endgroup$
    – Zerox
    Jan 14, 2022 at 17:35
  • $\begingroup$ @Zerox It is so. $\endgroup$
    – Mark Viola
    Jan 14, 2022 at 18:35

By the Stolz-Cesàro theorem (https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem), \begin{eqnarray} \lim_{n\to\infty}x_n&=&\lim_{n\to\infty}\frac1n \sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)\\ &=&\lim_{n\to\infty}\frac{\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)}{n}\\ &=&\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1} \left(\frac{n+2-k}{k}-\log(k)\right)-\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)}{(n+1)-n}\\ &=&\lim_{n\to\infty}\sum_{k=1}^{n+1} \left(\frac{n+1-k}{k}+\frac1k-\log(k)\right)-\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)\\ &=&\lim_{n\to\infty}\sum_{k=1}^{n+1}\frac1k+ \sum_{k=1}^{n+1}\left(\frac{n+1-k}{k}-\log(k)\right)-\sum_{k=1}^n \left(\frac{n+1-k}{k}-\log(k)\right)\\ &=&\lim_{n\to\infty}\sum_{k=1}^{n+1}\frac1k-\log(n+1)\\ &=&\gamma \end{eqnarray}


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