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I want to determine the convergence of the series and to do so, consider that

$$\sum_{n≥ 1}\left(n^{-1-\frac{1}{n}}\right)=\sum_{n≥ 1}\left(n^{\frac{-n-1}{n}}\right)$$

I consider at first glance that the series converges, and to determine this I was thinking of using the limit comparison test; however I would have to find a sequence $b_n$ such that $\frac{a_n}{b_n}=L≥0$ Clearly where $b_n$ converges.

Any help? or any other way to approach this? thanks.

Wrloord
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    note that $n^{1/n} \le 2, n \ge 2$ so $n^{-1-\frac{1}{n}} \ge 1/(2n), n \ge 2$ what do you conclude? – Conrad Mar 09 '22 at 22:43
  • I was thinking of using the basic comparison test, and since $\sum_{n≥2}\frac{1}{2n}$ diverges then the serie I am studying would also diverge, right? – Wrloord Mar 09 '22 at 22:58
  • If $a_n=n^{-1-\frac{1}{n}}$ then $$2^n a_{2^n}=2^{-\frac{n}{2^n}}\longrightarrow 1 $$ So, you series diverges by cauchy condensation. – Matthew H. Mar 10 '22 at 03:12

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You have $$ n^{-1-\frac{1}{n}} = \frac{1}{n\cdot n^{\frac{1}{n}}} = \frac{1}{n \cdot e^{\frac{\ln n}{n}}} $$ Now, what is the limit of $e^{\frac{\ln n}{n}}$, and can you use this to conclude?

Clement C.
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