# 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?

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 $\sigma > \frac{1}{2}$.

• Thanks for the detailed treatment! – Shree Apr 13 '18 at 4:29

$\frac1{n^\sigma(n^\sigma - 1)} \sim \frac1{n^{2\sigma}}$ Now when the last one converges ?

• Thanks. $\sum \frac{1}{n^{2 \sigma}} = \zeta(2\sigma)$ converges for $2\sigma > 1$ – Shree Apr 13 '18 at 4:27

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 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.