How to prove this sequence converges to 1?

Question

How can we prove this converges to 1?

$a_n = \sqrt[n]{\sum_{i=1}^n \frac{1}{\sqrt[i]{i} }} $

It is upper bounded by 1 because $ \frac{1}{\sqrt[n]{n}} < 1$ and $\sqrt[n]{n}$ converges to 1 but that's all I have.

Answer 1

Show that for $x\ge1$, $e^{-1/e}\le x^{-1/x}\le1$. Thus, $$ \left(ne^{-1/e}\right)^{1/n}\le\left(\sum_{i=1}^ni^{-1/i}\right)^{1/n}\le n^{1/n} $$

Answer 2

Observe that the sequence $n^{1/n}$ is eventually decreasing. Indeed, the function $x^{1/x}$, with $x\geq 1$, can be written as $e^{\frac 1x \log x}$, and taking the derivative we see $$ (x^{1/x})' = x^{1/x} ( -\frac {1}{x^2} \log x + \frac{1}{x^2} ) = x^{1/x} \frac{1}{x^2} (1 - \log x) < 0, \ \text{ if } x\geq 3. $$ Hence, we get $$ (n-2) \frac{1}{3^{1/3}} \leq \sum_{i=1}^n \frac{1}{i^{1/i}} \leq n. $$ Taking the $1/n$-th root on the last expression, we get that $a_n \to 1$ as the $1/n$-th power of both sides of the above inequality converges to $1$.