Is $\sum_{n=2}^\infty \frac{1}{n^\sigma \dot (n^\sigma-1)} $ convergent for $ 0 < \sigma < 1 , \sigma \in \mathbb{R} $ I am trying to figure out convergence or divergence of $\sum_{n=2}^\infty \frac{1}{n^\sigma \dot (n^\sigma-1)} $ for $ 0 \leq \sigma \leq 1 , \sigma \in \mathbb{R} $ we know the series is convergent for $ \sigma = 1 $ as it turns into a telescoping series. It is divergent at $ \sigma = 0 $. Where is the boundary? 
I have tried divergence test, ratio test, root test. Ratio year and root test give limit of 1 and all three tests are inconclusive  are inconclusive. 
For limit comparison test, I tried comparing with convergent telescoping series $\sum_{n=2}^\infty \frac{1}{n \dot (n-1)} $ unfortunately it’s not useful.
I couldn’t think of any function for integral test.
Any other suggestions? 
 A: $\frac1{n^\sigma(n^\sigma - 1)} \sim \frac1{n^{2\sigma}}$ Now when the last one converges ? 
A: Hint: We have that for $N$ sufficiently large
\begin{align}
\frac{1}{2}N^\sigma \leq N^{\sigma}-1\leq N^\sigma
\end{align}
A: It diverges for $\sigma \in [0,\frac{1}{2}]$ and converges for $\sigma \in (\frac{1}{2},1]$
 $$ \Sigma_{n=2}^{\infty} \frac{1}{n^{\sigma}(n^{\sigma} - 1)} > \Sigma_{n=2}^{\infty} \frac{1}{n^{2\sigma}}$$
 And $\Sigma_{n=1}^{\infty} \frac{1}{n^{2\sigma}}$ diverges for $\sigma \leq \frac{1}{2}$, thus the series diverges for $\sigma \leq \frac{1}{2}$. 
      $$ \Sigma_{n=2}^{\infty} \frac{1}{n^{\sigma}(n^{\sigma} - 1)} < \Sigma_{n=2}^{\infty} \frac{1}{(n^{\sigma} - 1)^2}$$
    For $n \geq \lceil 2^{\frac{1}{\sigma}}\rceil $, $ n^{\sigma} - 1 \geq \frac{n^{\sigma}}{2}$
    $\Rightarrow  \Sigma_{n=k}^{\infty} \frac{1}{(n^{\sigma} - 1)^2} \leq \Sigma_{n=k}^{\infty} \frac{4}{n^{2\sigma}}$ where $ k = \lceil 2^{\frac{1}{\sigma}}\rceil$.
    And $\Sigma_{n=k}^{\infty} \frac{1}{n^{2\sigma}}$ converges for $\sigma > \frac{1}{2}$, thus $\Sigma_{n=2}^{\infty} \frac{1}{n^{\sigma}(n^{\sigma} - 1)}$ converges for  $\sigma > \frac{1}{2}$.
A: A more compact approach would be to use limit comparison test. Compare $ a_n $ and $ b_n $ terms suggested by @Youem.
$ \frac{1}{n^{2\sigma}}$ and $ \frac{1}{n^\sigma(n^\sigma-1)} $. As the ratio of limits $$ \frac{\lim_{n\rightarrow\infty} \frac{1}{n^{2\sigma}}}{\lim_{n\rightarrow\infty} \frac{1}{n^\sigma(n^\sigma-1)}} = 1 $$
the rate of growth is comparable, thus the two series converge ($ 2\sigma > 1 $) or diverge ($ 2\sigma \leq 1 $) together.
