Find $\lim\limits_{n \to +\infty}\frac {n}{(\log(n+1))^n}$

Question

I'm studying the convergence of the series $$\sum_{n=1}^\infty \frac {n}{(\log(n+1))^n}$$

$\frac {n}{(\log(n+1))^n}>0, \forall n \ge 1$
throught the root test I can proof the series converges but I can't solve the limit

$\lim\limits_{n \to +\infty}\frac {n}{(\log(n+1))^n}$ should be $0$ .

If you can prove the series converges using the root test, then the limit is automatically $0$. — rtybase, Jan 22 '18 at 11:11
If you have only the limit, you can't ... but this is not what the question states. — rtybase, Jan 22 '18 at 11:14
@rtybase ok, I put it more clearly, now it is the question states. — Anne, Jan 22 '18 at 11:16
It's still almost the same ... so I can prove the series converges using ratio test. So the series converges, then the sequence has the limits $0$. It's a valid proof. — rtybase, Jan 22 '18 at 11:19
$\displaystyle 0\leq\lim\limits_{n\to\infty}\frac{n}{(\ln(n+1))^n} = \lim\limits_{n\to\infty}\frac{e^n-1}{n^{e^n-1}} \leq \lim\limits_{n\to\infty}\frac{e^n}{n^n} \leq \lim\limits_{n\to\infty}\frac{1}{2^n} = 0 $ — user90369, Jan 22 '18 at 11:42

rtybase · Answer 1 · 2018-01-22T11:43:15.403

I think ratio test might be easier $$\color{red}{0\leq}\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\rightarrow\infty} \frac{\frac {n+1}{(\log(n+2))^{n+1}}}{\frac {n}{(\log(n+1))^n}}=\\ \lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\left(\frac{\log(n+1)}{\log(n+2)}\right)^n\frac{1}{\log(n+2)}\leq ...$$ since $\log(n+1)< \log(n+2)$ thus $\left(\frac{\log(n+1)}{\log(n+2)}\right)^n<1$ $$...\leq\lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\frac{1}{\log(n+2)}= \lim\limits_{n\rightarrow\infty}\frac{n+1}{n}\cdot \lim\limits_{n\rightarrow\infty}\frac{1}{\log(n+2)}=\\ 1\cdot \lim\limits_{n\rightarrow\infty}\frac{1}{\log(n+2)} \color{red}{=0}$$ Thus the series converges and the $\lim\limits_{n \to \infty}\frac {n}{(\log(n+1))^n}=0$

score 1 · Accepted Answer · answered Jan 22 '18 at 11:21

Using the root test $$ \sqrt[1/n]{n/\ln(n+1)^n} = n^{1/n}/\ln(n+1). $$ For the guy above we have $$ n^{1/n} = e^{\ln(n^{1/n})} = e^{\ln(n)/n}, $$ so that we can study the limit $$ \lim_n \ln(n)/n \overset{Hopital}{=} \lim_n 1/n=0. $$ Combining all together, along with the continuity of $e^x$, we get $$ \limsup_n \sqrt[1/n]{n/\ln(n+1)^n} = \limsup_n e^{\ln(n)/n}/\ln(n+1) =0.$$ Note that a full argument would be "consider $f:\mathbb{R}^+\to \mathbb{R},x\mapsto \sqrt[1/x]{x/\ln(x+1)^n}$ etc.", you find that the limsup exists, so that the lim exists and it is $0$ being the function positive (by comparison). Therefore for any sequence $\{x_n\}_{n\in\mathbb{N}}\subset \mathbb{R},x_n \to \infty$, you have $f(x_n)\to 0$. Pick $x_n=n$ and you are done.

score 0 · Answer 3 · answered Jan 22 '18 at 16:06

0

Let

$$a_n= \frac {n}{(\log(n+1))^n}\to0$$

indeed by ratio test

$$\frac{a_{n+1}}{a_n}=\frac {n+1}{ (\log(n+2))^{n+1} } \frac {(\log(n+1))^n }{ n }=\frac{n+1}{n} \frac{1}{\log(n+2)} \left(\frac { \log(n+1) }{ \log(n+2) }\right)^n\to0$$

indeed

$$\frac{n+1}{n}\to1, \quad \frac{1}{\log(n+2)}\to 0, \quad 0< \left(\frac { \log(n+1) }{ \log(n+2) }\right)^n <1$$

answered Jan 22 '18 at 16:06

user

154,566

Didn't you accidentally copy my solution? :) – rtybase Jan 22 '18 at 17:49
1

Yes I did but with a different style :) – user Jan 22 '18 at 17:56

Find $\lim\limits_{n \to +\infty}\frac {n}{(\log(n+1))^n}$

3 Answers3