For what $x>0$ series $\sum_{n=1}^{\infty} x^{1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}...+\frac{1}{\sqrt n}}$ is convergent?

Question

For what $x>0$ series $\sum_{n=1}^{\infty} x^{1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}...+\frac{1}{\sqrt n}}$ is convergent?

By logarithmic test, $$ \lim_{n\rightarrow \infty}\left(n \log\frac{u_n}{u_{n+1}}\right)$$ is infinite for $0<x<1$. So for $x\in (0,1)$, series is convergent. Am I correct? Thanks.

Answer 1

Since the sum $\displaystyle\sum_{k=1}^n\,\frac{1}{\sqrt{k}}$ is well approximated by $2\sqrt{n}$, the required series is convergent iff the series $$\sum_{n=1}^\infty\,x^{2\sqrt{n}}$$ is convergent. Since $$\int_0^\infty\,\exp\big(-\alpha\sqrt{t})\,\text{d}t=\frac{2}{\alpha^2}<\infty$$ for all $\alpha>0$, the required series is convergent for all $x\in(0,1)$ by the Integral Test. Surely, if $x\geq 1$, the required series is divergent. Ergo, you are correct.

Answer 2

By comparing with an integral, you can see that the exponent $$\sum_{k=1}^n \frac 1 {\sqrt k} \approx 2\sqrt{n}$$ so the series converges iff $|x|<1$.

Answer 3

Yes you are true.$$$$ First if $x\ge1$ then $x^{1+\cdots+\frac{1}{\sqrt n}}\ge x\ge1=>$ sum divergent.$$$$ Second if $0<x<1$. $$ \\1+\frac{1}{\sqrt2}+\cdots+\frac{1}{\sqrt{n}}\ge \frac{1}{\sqrt{n}}+\cdots+\frac{1}{\sqrt{n}}=\frac{n}{\sqrt{n}}=\sqrt n=> \\x^{1+\cdots+\frac{1}{\sqrt{n}}}\le x^{\sqrt{n}} \\\lim_{n\to\infty}\frac{\ln\frac{1}{x^\sqrt n}}{\ln n}=\lim_{n\to\infty}\frac{\sqrt{n}}{\ln n}\cdot \ln (\frac{1}{x})>1=> $$ sum convergent. https://ru.wikipedia.org/wiki/Логарифмический_признак_сходимости