$\sum_{n=1}^{\infty }\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)} $ converges or diverges? I am trying to determine whether this series converges or diverges: $\sum_{n=1}^{\infty }\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)}$.
Here is my solution: I called: $a_{n}=\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)}$. Then, I used the root test as follows: $\lim_{n \to \infty  }\left | a_{n} \right |^{\frac{1}{n}}=\lim_{n \to \infty}\left(\frac{2n+5}{7n+6}\right)^{\log(n+1)}$. Then I called $x_{n}=(\frac{2n+5}{7n+6})^{\log(n+1)}$, Instead of computing $\lim_{n \to \infty}x_{n}$, I computed first $$\lim_{n \to \infty}\log(x_{n})=\lim_{n \to \infty}\log(n)\log\left(\frac{2n+5}{7n+6}\right)=\log(\frac{2}{7})\lim_{n \to \infty}\log(n)=-\infty,$$ therefore: $\lim_{n \to \infty}x_{n}=\lim_{n \to \infty}e^{\log(x_{n})}=0$. Therefore: $\lim_{n \to \infty  }\left | a_{n} \right |^{\frac{1}{n}}=0< 1$. So by the root test, the series converges.
Can you please let me know whether my solution is correct (especially the last steps) or not? if there is a mistake, please let me know how I should fix it. Also, if you are aware of a better way of solving the problem , please do let me know. Thanks!
 A: A bit more complicated than I thought:
$$\left (\frac{2 n+5}{7 n+6}\right)^{n \log{(n+1)}} \sim \left (\frac{2 }{7}\right)^{n \log{n}} \frac{\left(1+\frac{5/2}{n}\right)^{n \log{n}}}{\left(1+\frac{6/7}{n}\right)^{n \log{n}}}$$
This becomes
$$\left (\frac{2 }{7}\right)^{n \log{n}} n^{23/14} \quad (n \to \infty)$$
So then the question becomes, does
$$\sum_{n=1}^{\infty} \left (\frac{2 }{7}\right)^{n \log{n} } n^{23/14}$$
converge?  The answer is yes, by the integral test, because
$$\int_1^{\infty} dt \, t^{\alpha} \, e^{- \beta t}$$
converges for any positive $\alpha$ and $\beta$, and because $n \lt n \log{n}$.  (Here, $\alpha = 23/14$ and $\beta = \log{(7/2)}$.)
A: Another way to to prove the convergence would be to study the integral
$$\int_0^\infty \left(\frac{2x+5}{7x+6} \right)^{x\ln x}\, dx.$$
The integrand is a continuous function in $x\to 0$ and on infinity is majorated by $c(2/7)^x$ for some positive constant $c$, hence the integral converges; by an integral criterion, so does the series.
Your method works, too.
A: As for $n \geq 3$ we have  $$\frac{2n+5}{7n+6} < \frac{3}{7} <1$$
we get for $n \geq 3$
$$\left(\frac{2n+5}{7n+6}\right)^{n\log(n+1)} < \left(\frac{3}{7}\right)^{n\log(n+1)}< \left(\frac{3}{7}\right)^{n}$$
Thus, by comparison to a geometric series, the series is convergent.
