I am trying to prove convergence of following series using Taylor series.
$$\sum_{n=1}^{\infty}\left(e^{\frac{1}{\sqrt{n}}}-1-\frac{1}{\sqrt{n}} - \frac{1}{2n}\right)$$
$$e^{\frac{1}{\sqrt{n}}}=1+\frac{1}{\sqrt{n}}+\frac{1}{2n} +\left(\frac{1}{\sqrt{n}}\right)^2 r\left(\frac{1}{\sqrt{n}}\right)$$
When $\frac{1}{\sqrt{n}}\to 0$ then $r\left(\frac{1}{\sqrt{n}}\right)\to 0$. So if $n\to\infty$ then $r\left(\frac{1}{\sqrt{n}}\right)\to 0$ $$\sum_{n=1}^{\infty}\frac{1}{n^2}+r\left(\frac{1}{\sqrt{n}}\right)=\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=1}^{\infty}r\left(\frac{1}{\sqrt{n}}\right)$$
I am not sure about convergence of $\sum_{n=1}^{\infty}r\left(\frac{1}{\sqrt{n}}\right)$. I can't show it. Could you help me ?
Hint. One may write, by the Taylor series expansion, as $n \to \infty$, $$ e^{\large\frac1{\sqrt{n}}}=1+\frac{1}{\sqrt{n}}+\frac{1}{2n}+\frac{1}{6n^{3/2}} +\left(\frac{1}{\sqrt{n}}\right)^3 r\left(\frac{1}{\sqrt{n}}\right) $$ with $r(u) \to 0$ as $u \to 0$, giving that, there exists some $n_0>0$ such that $\displaystyle \left|r\left(\frac{1}{\sqrt{n}}\right)\right|\le1$ for all $n \ge n_0$, that is $$ 0 \le\left(\frac{1}{\sqrt{n}}\right)^3 r\left(\frac{1}{\sqrt{n}}\right)\le\frac{1}{n^{3/2}},\qquad \qquad n\ge n_0, $$ giving the convergence of $$ \sum_{n=1}^{\infty}\left(e^{\large\frac1{\sqrt{n}}}-1-\frac{1}{\sqrt{n}}-\frac{1}{2n}\right)=\sum_{n=1}^{\infty}\frac{1}{6n^{3/2}}+\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}\right)^3 r\left(\frac{1}{\sqrt{n}}\right) $$by comparison to a convergent $p$-series.