how to find the value of $\alpha $ the given series is converge .. find the  value  of $\alpha $ the given  series  is  converge ..
$a)$ $\sum _{n=1}^{\infty}(\sqrt [n] n - 1)^{\alpha} $,
$b)$ $\sum_{n=1}^{\infty} ((1+\frac{1}{n})^{n+1}  - e) ^{\alpha}$
My attempts  : For  $a)$  i know that $\frac {logn}{n}  < (\sqrt [n] n -1 )$
Now $\sum_(\frac {logn}{n})^{\alpha}  < \sum (\sqrt [n] n -1 )^\alpha$ is converge  if  $\alpha  > 1$
for $ b)$ im very confused   i don't  know  how  to  starts,
Any Hints/solution  will  be  appreciated 
thanks u 
 A: For the first one we have that for $\alpha \le 0$ the series diverges since $a_n \not \to 0$ therefore assuming $\alpha >0$ we have
$$\sqrt [n] n=e^{\frac{\log n}{n}}=1+\frac{\log n}{n}+O\left(\frac{\log^2 n}{n^2}\right)$$
therefore
$$(\sqrt [n] n - 1)^{\alpha}=\left(\frac{\log n}{n}+O\left(\frac{\log^2 n}{n^2}\right)\right)^{\alpha}$$
which converges for $\alpha>1$ and diverges for $\alpha\le1$ by limit comparison test with $\left(\frac{\log n}{n}\right)^{\alpha}$.
For the second one we have also that for $\alpha \le 0$ the series diverges since $a_n \not \to 0$ therefore assuming $\alpha >0$ we have
$$\left(1+\frac{1}{n}\right)^{n+1}=e^{(n+1)\log \left(1+\frac{1}{n}\right)}=e^{(n+1)\left(\frac{1}{n}-\frac{1}{2n^2}+O(1/n^3)\right)}=e^{\left(1+\frac{1}{2n}+O(1/n^2)\right)}=e\left(1+\frac{1}{2n}+O(1/n^2)\right)=e+\frac{e}{2n}+O(1/n^2)$$
and therefore
$$ \left(\left(1+\frac{1}{n}\right)^{n+1}  - e\right) ^{\alpha}= \left(\frac{e}{2n}+O(1/n^2)\right) ^{\alpha}$$
which converges for $\alpha>1$ and diverges for $\alpha\le1$ by limit comparison test with $\left(\frac{1}{n}\right)^{\alpha}$.
