Asymptotic behavior of $\sum\limits_{k=1}^{n}e^{1/\sqrt k}$

Question

Is where any way to find a function $f(n)$ such that $\sum\limits_{k=1}^{n}e^{1/\sqrt k} = f(n) + O(1)$?

Answer 1

$$ e^{\frac{1}{\sqrt{k}}}=1+\frac{1}{\sqrt{k}}+\frac{1}{2k}+\mathcal{O}\left(\frac{1}{k^{3/2}}\right) $$ Thus $$ \sum_{k=1}^n{e^{\frac{1}{\sqrt{k}}}}=n+\sum_{k=1}^n{\frac{1}{\sqrt{k}}}+\frac{1}{2}\sum_{k=1}^n{\frac{1}{k}}+\sum_{k=1}^n{\mathcal{O}\left(\frac{1}{k^{3/2}}\right)} $$ However, $$ \sum_{k=1}^n{\frac{1}{k}}=\ln n+\mathcal{O}(1) $$ Since $\sum{\frac{1}{k^{3/2}}}$ converges, you have $$ \sum_{k=1}^n{\mathcal{O}\left(\frac{1}{k^{3/2}}\right)}=\mathcal{O}(1) $$ and $$ \sum_{k=1}^n{\frac{1}{\sqrt{k}}}=\sqrt{n}\times\frac{1}{n}\sum_{k=1}^n{\sqrt{\frac{n}{k}}}=\sqrt{n}\left(\int_0^1{\frac{dx}{\sqrt{x}}}+o(1)\right)=2\sqrt{n}+o(\sqrt{n}) $$ The $o(\sqrt{n})$ is not necessary a $\mathcal{O}(1)$, we have to do better. Let $f(x)=\frac{1}{\sqrt{x}}$ and $$ u_n=\sum_{k=1}^n{f(k)}-\int_0^n{f(x)dx} $$ $$ v_n=\sum_{k=1}^{n-1}{f(k)}-\int_0^n{f(x)dx} $$ First, $u_n-v_n=f(n)=\frac{1}{\sqrt{n}}\underset{n\rightarrow +\infty}{\longrightarrow}0$, moreover since $f$ is decreasing you have $$ u_{n+1}-u_n=f(n+1)-\int_{n}^{n+1}{f(x)dx}\leqslant 0 $$ thus $(u_n)$ is decreasing and by the same argument, $(v_n)$ is increasing. Finally, the sequences $(u_n)$ and $(v_n)$ are adjacents and they converge toward the same limit $\ell$. Thus $u_n=\ell+o(1)$ and $$ \sum_{k=1}^n{\frac{1}{\sqrt{k}}}=2\sqrt{n}+\ell+o(1)=2\sqrt{n}+\mathcal{O}(1) $$ You finally have that $$ \sum_{k=1}^n{e^{\frac{1}{\sqrt{k}}}}=n+2\sqrt{n}+\frac{1}{2}\ln n+\mathcal{O}(1) $$

Answer 2

$e^{1/\sqrt k} \ge 1$ for all $k$, hence the series $\sum\limits_{k=1}^{ \infty}e^{1/\sqrt k}$ is divergent, which means that $\sum\limits_{k=1}^{n}e^{1/\sqrt k} \to \infty$ as $n \to \infty.$

Consequence: the sequence $(\sum\limits_{k=1}^{n}e^{1/\sqrt k})$ is unbounded.

Answer 3

Expand $exp(x)=1+x+x^2/2+O(x^3)$, and find sum(1), sum(1/sqrt(k)) and sum(1/k)