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Let $(a_n) = \left(1 - \frac{1}{n^{1/n}}\right)$.

Now consider the series $\, S = \sum_{n=0}^\infty |a_n|^p,\, 1 \leq p < \infty, \, p \in \mathbb{R}$. I am interested to know whether the series $S$ diverges for all $p$ or is there some $p$ for which the series converges. I am not able to prove that it diverges for all $p$ but I think that would be the case. I tried some of the basic convergence tests I know such as comparison test, ratio test, root test, integral test, Cauchy's condensation test.

Can someone help out here ?

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  • $\begingroup$ isn't the term undefined for $n=0$? $\endgroup$
    – person
    Mar 7, 2022 at 19:45
  • $\begingroup$ The sequence is assume to start from $1$ that is $n = 1$ $\endgroup$ Mar 8, 2022 at 2:11

2 Answers 2

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At least for $p > 2$ the series is convergent: Set $c_n:=n^{1/n}-1$. For $n\ge 2$ we have $$ n=(c_n+1)^n=\sum_{k=0}^n {n \choose k} c_n^k \ge {n \choose 2}c_n^2=\frac{n(n-1)}{2}c_n^2, $$ hence $$ 0 \le c_n \le \frac{\sqrt{2}}{\sqrt{n-1}}, $$ so $$ 0 \le a_n=\frac{c_n}{n^{1/n}} \le c_n \le \frac{\sqrt{2}}{\sqrt{n-1}}. $$ For $p>2$ this yields a convergent majorant: $$ a_n^p \le \frac{2^{p/2}}{(n-1)^{p/2}}. $$

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$\begin{array}\\ a_n &= 1 - \frac{1}{n^{1/n}}\\ &= 1 - n^{-1/n}\\ &= 1 - e^{-\ln(n)/n}\\ &= 1 - (1-\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2}))\\ &= \frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})\\ &= \frac{\ln(n)}{n}(1+O(\frac{\ln(n)}{n}))\\ a_n^p &= \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))^p\\ &= \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))\\ \sum_{n=1}^{\infty} a_n^p &= \sum_{n=1}^{\infty} \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))\\ \end{array} $

This converges for $p > 1$ since $\ln(n) =o(n^c) $ for any $c > 0$ (choose $c$ such that $cp < p-1$ such as $c = (p-1)/(2p)$ so that $\frac{\ln^p(n)}{n^p} =o(\frac{n^{(p-1)/2}}{n^p}) =o(\frac1{n^{(p+1)/2}}) $ and $(p+1)/2 > 1$) and diverges for $p \le 1$ by comparison with $\sum \frac1{n} $.

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