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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?

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

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  • $\begingroup$ Thanks for the detailed treatment! $\endgroup$ – Shree Apr 13 '18 at 4:29
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$\frac1{n^\sigma(n^\sigma - 1)} \sim \frac1{n^{2\sigma}}$ Now when the last one converges ?

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  • $\begingroup$ Thanks. $ \sum \frac{1}{n^{2 \sigma}} = \zeta(2\sigma)$ converges for $ 2\sigma > 1 $ $\endgroup$ – Shree Apr 13 '18 at 4:27
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Hint: We have that for $N$ sufficiently large \begin{align} \frac{1}{2}N^\sigma \leq N^{\sigma}-1\leq N^\sigma \end{align}

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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.

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