1
$\begingroup$

I need help with the following exercise that reads as follows

Calculate the limit of the sequence defined by:

$$x_n=\frac{1}{n}(n+\frac{n-1}{2}+\frac{n-2}{3}+...+\frac{2}{n-1}+\frac{1}{n}-\log(n!))$$

The idea is to find this limit by means of asymptotic equivalences, for this purpose consider the following

As $\log(n!)∼n\log(n)$ then $x_n ∼ \frac{1}{n}(n+\frac{n-1}{2}+\frac{n-2}{3}+...+\frac{2}{n-1}+\frac{1}{n}-n\log(n))$

I thought about establishing this equivalence, however I do not know how to arrive at the value of the limit.

Thank you for your help

$\endgroup$
2
  • $\begingroup$ Calculations indicate that the limit is $\gamma$ $\endgroup$
    – Peter
    Commented Jan 8, 2022 at 14:08
  • 1
    $\begingroup$ Actually: $\ln(n!) \sim n\ln(n) - n$. $\endgroup$
    – Enrico M.
    Commented Jan 8, 2022 at 15:12

1 Answer 1

Reset to default
2
$\begingroup$

Hint. Note that $$n+\frac{n-1}{2}+\frac{n-2}{3}+...+\frac{2}{n-1}+\frac{1}{n}=\sum_{k=1}^n\frac{n+1-k}{k}=(n+1)H_n-n$$ where $H_n=\sum_{k=1}^n\frac{1}{k}$ is the $n$-th harmonic number.

Recall that $H_n=\log(n)+\gamma +o(1)$. Moreover, use Stirling approximation for $n!$.

$\endgroup$
5
  • $\begingroup$ In other words, I can restate the succession as follows $\frac{1}{n}(\sum_{k=1}^n\frac{n+1-k}{k}-log(\sqrt{2πn}(\frac{n}{e})^n)$ ? or am I misreading the situation, it's honestly not clear to me. $\endgroup$
    – Wrloord
    Commented Jan 8, 2022 at 18:40
  • $\begingroup$ Yes, you also need $H_n=\log(n)+\gamma +o(1)$ $\endgroup$
    – Robert Z
    Commented Jan 8, 2022 at 19:38
  • $\begingroup$ Of course, now, I have to determine the limit of $x_n=\frac{1}{n}(\sum_{k=1}^n\frac{n+1-k}{k}-log(\sqrt{2πn}(\frac{n}{e})^n)$ ?, but it is not clear to me how to involve that $H_n=log(n)+γ+o(1)$ Could you give me a hint on how to start calculating the limit? $\endgroup$
    – Wrloord
    Commented Jan 8, 2022 at 19:41
  • $\begingroup$ Well, however, I have not found the way to find the limit from the last expression, do you think you could help me a little more? I would appreciate it very much. $\endgroup$
    – Wrloord
    Commented Jan 8, 2022 at 22:07
  • 1
    $\begingroup$ It remains to evaluate the limit of $\frac{1}{n}((n+1)(\log(n)+\gamma)-n-(n\log(n) - n))$ $\endgroup$
    – Robert Z
    Commented Jan 9, 2022 at 9:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .