<s>[INST] Let $a_n=2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt{k}}$. Show $a_n$ converges and $1<\lim_{n\to\infty}a_n<2$. I am able to show that $\lim_{n\to\infty}a_n\le 2$, but I can't think of a way to show strict inequality.
Here's my work so far:
By Riemann sums,
$$\int_1^{n+1}\frac{1}{\sqrt x}\,dx<\sum_{k=1}^n\frac{1}{\sqrt k}<\int_0^n\frac{1}{\sqrt x}\,dx.$$
$$2\sqrt{n+1}-2<\sum_{k=1}^n\frac{1}{\sqrt k}<2\sqrt n$$
So,
$$0<2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}<2\sqrt n+2-2\sqrt{n+1},$$
which implies,
$$0<2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}<2\quad\forall\,n\ge1.$$
$$\begin{align}
a_{n+1}-a_n & =\left(2\sqrt{n+1}-\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\right)-\left(2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}\right)\\
& =2(\sqrt{n+1}-\sqrt n)-\frac{1}{\sqrt{n+1}}\\
&=\frac{2}{\sqrt{n+1}+\sqrt n}-\frac{1}{\sqrt{n+1}}\\
&=\frac{\sqrt{n+1}-\sqrt n}{\sqrt{n+1}(\sqrt{n+1}+\sqrt n)}>0.
\end{align}$$
Since, $a_1=1$, $\{a_n\}$ is a monotonically increasing sequence, 
and 
$$1<a_n<2,\quad\forall\,n\ge1.$$
So, $\{a_n\}$ converges, and 
$$1<\lim_{n\to\infty}a_n\le2.$$
 [/INST] You just have to make your initial estimate more quantitative:
\begin{eqnarray*}
2\sqrt{n+1}-2-\sum_{m=1}^{n}\frac{1}{\sqrt{m}} & = & \int_{1}^{n+1}\frac{1}{\sqrt{x}}\,{\rm d}x-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}\\
 & = & \sum_{m=1}^{n}\int_{m}^{m+1}\underbrace{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{m}}}_{\leq0}\,{\rm d}x\\
 & \leq & \int_{1}^{2}\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{1}}\,{\rm d}x\\
 & = & 2\sqrt{2}-3<0
\end{eqnarray*}
and hence
\begin{eqnarray*}
a_{n}-2 & = & 2\sqrt{n}-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}-2\\
 & = & 2\sqrt{n+1}-2-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}+2\left(\sqrt{n}-\sqrt{n+1}\right)\\
 & \leq & 2\sqrt{2}-3+2\left(\sqrt{n}-\sqrt{n+1}\right)\\
 & \leq & 2\sqrt{2}-3.
\end{eqnarray*}
This implies
$$
\lim_{n\to\infty}a_{n}\leq2+2\sqrt{2}-3<2.
$$
</s>