0
$\begingroup$

Test the convergence of the series whose nth term is given using Root test $$a_{n}=\left({\frac{\log n}{log(n+1)}}\right)^{n^2\log n}$$

I know that if $lim_{n \rightarrow \infty}(a_{n})^{1/n}\lt 1$,the series converges

So, let $$ lim_{n \rightarrow \infty}\left({\frac{\log n}{log(n+1)}}\right)^{n\log n}=y$$ Taking log on both sides,

$$lim_{n \rightarrow \infty}n\log n\log\left({\frac{\log n}{log(n+1)}}\right)=\log y$$ But this is leading to complicated expressions.Is there a simpler way to approach this?

$\endgroup$
1
$\begingroup$

Note that

$$\left(\frac{\log n}{\log(n+1)}\right)^{n\log n}=\left(\frac{\log n}{\log n+\log(1+\frac1n)}\right)^{n\log n}=\left(\frac{\log n^n}{\log n^n+\log(1+\frac1n)^n}\right)^{\log n^n}=\left(\frac{1}{1+\frac{\log(1+\frac1n)^n}{\log n^n}}\right)^{\log n^n}\to\frac1e$$

indeed

$$\left(1+\frac{\log(1+\frac1n)^n}{\log n^n}\right)^{\log n^n}=e^{\log n^n\cdot \log \left(1+\frac{\log(1+\frac1n)^n}{\log n^n}\right)}=e^{\log n^n\cdot \left(\frac{\log(1+\frac1n)^n}{\log n^n}+o\left(\frac{1}{logn^n}\right)\right)}=e^{\left(\log(1+\frac1n)^n+o(1)\right)}\to e$$

$\endgroup$
1
$\begingroup$

Hint

When $n$ is large $$\frac{\log (n)}{\log (n+1)}\sim 1-\frac 1 {n \log(n)}$$ $$\log \left(\frac{\log (n)}{\log (n+1)}\right)\sim -\frac 1 {n \log(n)}$$

$\endgroup$
0
$\begingroup$

I think you can construct an inequality first to simplify the expression:

Because $0<\frac{log(n)}{log(n+1)}<1 \quad$ and $\quad n^2\log(n)\ge n >1$ $\qquad$ $\forall n\in\mathbb N, n>1$

We have $$a_{n}=\left({\frac{\log n}{log(n+1)}}\right)^{n^2\log n}<\left({\frac{\log n}{log(n+1)}}\right)^{n^2}=b_{n}$$

We now apply root test to $\sum b_{n}$ and evaluate: $$ lim_{n \rightarrow \infty}\left({\frac{\log n}{log(n+1)}}\right)^{n}=1$$

We get an indeterminate result. But, we can use Raabe's test as an extension: $$ lim_{n \rightarrow \infty}\left(n\cdot\bigr({\frac{b_{n}}{b_{n+1}}}\bigr)^{n}-1\right)=\infty$$ This means $\sum b_{n}$ converges. Thus, the original series $\sum a_{n}$ converges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.