An exercise on series of function convergence $\sum_{n=1}^{\infty} \frac{(n+1)^x}{n^2 \ln n}$. I am asked to show the various tipes of convergence of this series:
$$\sum_{n=1}^{\infty} \frac{(n+1)^x}{n^2 \ln n}$$ where $x \in R$.
I notice that for  $x \ge 1$ the series diverges because $ \frac{n}{n^2 \ln n} < \frac{n+1}{n^2 \ln n}$ and $\frac{1}{n \ln n}$ diverges from the integral test.
Anyone mind helping me with $x < 1$? I tried root and ratio test.  
 A: If $x<1$, we have, as $n \to \infty$,
$$
\frac{(n+1)^x}{n^2 \ln n}=\frac{n^x(1+1/n)^x}{n^2 \ln n} \sim \frac{1}{n^{2-x} \ln n}
$$ 
then by applying for example the Cauchy condensation test to the latter series you conclude that your initial series is convergent.

Edit. One may observe that


*

*if $0\leq x<1$, then as $n \to \infty$,  $$ \frac{1}{n^{2-x} \ln n} \leq
   \frac{(n+1)^x}{n^2 \ln n}=\frac{n^x(1+1/n)^x}{n^2 \ln n} \leq
   \frac{e^x}{n^{2-x} \ln n} $$

*if $x<0$, then as $n \to \infty$,$$ 0<   
   \frac{(n+1)^x}{n^2 \ln n}=\frac{1}{n^2(n+1)^{|x|} \ln n} \leq   
   \frac1{n^2 \ln n} $$


both cases giving the announced convergence.
A: Start the summation from $n=2$, because $\ln 1=0$ (just to be precise). Now, to the point, for $x<1$ $$\frac{(n+1)^x}{n^2\ln n}\sim_{\infty}\frac{n^x}{n^{2}\ln n}=\frac{1}{n^{2-x} \ln n}<\frac{1}{n^a}$$ with $a>1$ which converges.
A: As this is a series with positive terms, you can use equivalents:
$$\frac{(n+1)^x}{n^2\log n}\sim_\infty\frac{n^x}{n^2\log n}=\frac{1}{n^{2-x}\log n}. $$
This is a Bertrand's series, i. e. of type $\dfrac1{n^a\log^bn}$ and it is known to converge if and only $a>1$ or ($a=1$ et $b>1$) (the latter case is proved with the integral test).
Thus, in the present case, the series converges if and only if $2-x>1$, i.e. $x<1$.
If you're not allowed to use equivalence, you may adapt the idea behind equivalence as follows:
$$\frac{(n+1)^x}{n^2\log n}=\Bigl(1+\frac1n\Bigr)^x\frac{1}{n^{2-x}\log n}<2^x\frac{1}{n^{2-x}\log n},$$
which is proportional to a Bertrand's series &c.
