To test convergence of infinite series $x^2(\log 2)^p + x^3(\log 3)^p + x^4(\log 4)^p +\dots$ 
To test convergence of infinite series $x^2(\log 2)^p + x^3(\log 3)^p + x^4(\log 4)^p +\dots$

My approach to the above problem:
let $u_n = x^{(n+1)}(\log(n+1))^p$. the $u_{(n+1)}= x^{(n+2)}(\log(n+2))^p$
now, $n \log\frac{u_n}{u_{n+1}}$=n\log($\frac{1}{x}$) + n$p$[\log(\log(n(1+$\frac{1}{n}$))-(\log(n(1+$\frac{2}{n}$)))]. After that I have used the expansion for $\log (1+x)$ and got stuck. 
It seems if I take the limit of $n \log\frac{u_n}{u_{n+1}}$, i.e $\lim_{n\to\infty}n \log\frac{u_n}{u_{n+1}}$ =$\infty$ $(>1)
$$\Rightarrow$The series seems to be convergent, which is not the answer. Please clarify the mistake and guide me.
 A: Testing the infinite series:
$$x^2\ln^{\text{p}}(2)+x^3\ln^{\text{p}}(3)+x^4\ln^{\text{p}}(4)+\dots=\sum_{n=2}^{\infty}x^n\ln^{\text{p}}(n)$$
Use the ratio test:
$$\lim_{n\to\infty}\left|\frac{x^{n+1}\ln^{\text{p}}(n+1)}{x^n\ln^{\text{p}}(n)}\right|=\lim_{n\to\infty}\left|x\cdot\frac{\ln^{\text{p}}(n+1)}{\ln^{\text{p}}(n)}\right|=|x|\cdot\lim_{n\to\infty}\left|\frac{\ln^{\text{p}}(n+1)}{\ln^{\text{p}}(n)}\right|=|x|$$
By the ratio test, the series converges when $|x|<1$.

For, the limit:
$$\lim_{n\to\infty}\left|\frac{\ln^{\text{p}}(n+1)}{\ln^{\text{p}}(n)}\right|=\left[\lim_{n\to\infty}\left|\frac{\ln(n+1)}{\ln(n)}\right|\right]^{\text{p}}=\left[\lim_{n\to\infty}\left|\frac{\frac{1}{1+n}}{\frac{1}{n}}\right|\right]^{\text{p}}=\left[\lim_{n\to\infty}\left|\frac{1}{1+\frac{1}{n}}\right|\right]^{\text{p}}=1$$
A: The point is that the coefficient $(\log n)^p$ grows very slowly ($p$ being fixed). Therefore the series should have the same radius of convergence as the geometric series $x+x^2+x^3+\ldots$, something easily established using the ratio test. Using the expansion of log is not the right approach here.
A: There are a lot of possibilties to check convergence. 
E.g. the ratio test here for $p\ge 0$ is:
$$|\frac{x^{n+1}(\ln(n+1))^p}{x^n(\ln n)^p}|=|x|(\frac{\ln(n+1)}{\ln n})^p<1$$ 
This means $|x|<(\frac{\ln n}{\ln(n+1)})^p<1$: Convergence for $|x|<1$.
