# Does this series converge for any $1 \leq p < \infty$, $p \in \mathbb{R}$

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 ?

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

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: $$a_n^p \le \frac{2^{p/2}}{(n-1)^{p/2}}.$$
$$\begin{array}\\ a_n &= 1 - \frac{1}{n^{1/n}}\\ &= 1 - n^{-1/n}\\ &= 1 - e^{-\ln(n)/n}\\ &= 1 - (1-\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2}))\\ &= \frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})\\ &= \frac{\ln(n)}{n}(1+O(\frac{\ln(n)}{n}))\\ a_n^p &= \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))^p\\ &= \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))\\ \sum_{n=1}^{\infty} a_n^p &= \sum_{n=1}^{\infty} \frac{\ln^p(n)}{n^p}(1+O(\frac{\ln(n)}{n}))\\ \end{array}$$
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 $$\sum \frac1{n}$$.