Convergence of a sequence (possibly Riemann sum) Let $a_1, a_2, a_3, . . . , a_n$ be the sequence defined by 
$$
a_n = 2\sqrt{n}-\sum_{k=1}^{n}\frac{1}{\sqrt{k}} = 2\sqrt{n} - \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}-...-\frac{1}{\sqrt{n}}
$$
show that the sequence $a_n$ is convergent to some limit L, and that $1<L<2$. 

I tried looking at this as a Riemann sum. However, I failed to covert it to that. Any hints on that or alternate solution? Thanks
 A: Here's an elementary solution that uses calculus and one fact from analysis -- a bounded increasing sequence converges. 
First, I claim that $a_n$ is strictly increasing. To see why, calculate the difference $a_{n+1}-a_n$:
\begin{align}
a_{n+1}-a_n &= 2\sqrt{n+1}-2\sqrt{n} - \frac{1}{\sqrt{n+1}} \\
&= 2(\sqrt{n+1}-\sqrt n)\cdot\frac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1} + \sqrt n}
- \frac{1}{\sqrt{n+1}} \\
&= \frac{2}{\sqrt{n+1}+\sqrt n}-\frac{1}{\sqrt{n+1}} > 0.
\end{align}
The last inequality follows from the inequality $\frac{1}{2}(\sqrt{n+1}+\sqrt n)<\sqrt{n+1}$. It follows that $a_n$ is strictly increasing, and we conclude that $a_n>1=a_1$ for $n>1$.
Next, we're going to put upper and lower asymptotic bounds on $a_n$, as follows.
Letting $S_n=\sum_{k=1}^n k^{-1/2}$, I claim that
$$
2\sqrt{n+1} - 2 < S_n < 2\sqrt n.
$$
Proving this inequality will actually finish the problem: if we
subtract $2\sqrt n$ from all parts of the inequality and multiply by $-1$, we get
$$
2-2(\sqrt{n+1}-\sqrt n) > a_n > 0.
$$
This means that $a_n$ is bounded above, and therefore converges because we showed that it is strictly increasing. Taking the limit as $n\to\infty$ and writing $L=\lim_{n\to\infty} a_n$, we see that $2\geq L$. Then $1 < L \leq 2$. (With a finer analysis, you may be able to replace the $\leq$ with a $<$. My solution is not entirely satisfactory to that extent: can anyone offer an improvement?)
One way to find these bounds is to approximate the sum $S_n$ by integrals; a second way is to cleverly bound $k^{-1/2}$. I'll show you the first approach since it's probably what the question-writer wanted, and I'll show you the second because it's so slick.
To derive the bounds with integrals, start with the following integral inequality:
$$
\int_0^n x^{-1/2}\,dx < \sum_{k=1}^n k^{-1/2} < \int_1^{n+1} x^{-1/2}\,dx.
$$
If you haven't seen this sort of thing before, let me know and I'll be happy to explain. You can understand why this works by drawing the right picture. In any case, if you evaluate these integrals you'll produce the bound on $S_n$.
To derive the bounds with cleverness, start with the following square root inequality:
$$
\frac{2}{\sqrt{k+1}+\sqrt{k}} < \frac{1}{\sqrt k} < \frac{2}{\sqrt{k} + \sqrt{k-1}}.
$$
Completing the square on the left and right transforms the inequality into
$$
2(\sqrt{k+1}-\sqrt{k}) < \frac{1}{\sqrt k} < 2(\sqrt{k}-\sqrt{k-1}).
$$
Summing this inequality from $k=1$ to $k=n$ -- it telescopes -- yields
$$
2\sqrt{n+1} - 2 < S_n < 2\sqrt n,
$$
as desired.
A: This is maybe is little off-topic, but we can prove convergence and give an expression for the limit at the same time.
Let us note $\delta_n = a_n - a_{n-1}
=2\sqrt{n} - 2\sqrt{n-1} - \frac 1{\sqrt{n}}$.
Now use the fact that $\sqrt{}$ is analytic on $[0,1]$:
$$
\sqrt{1-z} = \sum_{m=0}^\infty \frac{(2m)!}{(m!)^2(1-2m)}
\left(\frac z{4}\right)^m
\\
\sqrt{n-1} - \sqrt{n} = \sqrt{n}\left(\sqrt{1-\frac 1n}-1\right) 
=\sqrt{n} \sum_{m=1}^\infty \frac{(2m)!}{(m!)^2(1-2m)}
\left(\frac 1{4n}\right)^m
\\
\delta_n = -\frac 1{\sqrt{n}} -2(\sqrt{n-1} - \sqrt{n})
=2\sqrt{n} \sum_{m=2}^\infty \frac{(2m)!}{(m!)^2(2m-1)}
\left(\frac 1{4n}\right)^m.
$$ 
Now everything is positive and we can use the Fubini–Tonelli theorem:
$$
\sum_{n=1}^\infty \delta_n =
2 \sum_{m=2}^\infty \frac{(2m)!}{(m!)^2(2m-1)4^m}
\zeta\left(m-\frac12\right)<\infty,
$$
because $\zeta\left(m-\frac12\right)$ is decreasing and hence bounded, and for the remaining series 
the Stirling formula yields:
$$
\frac{(2m)!}{4^m(m!)^2(2m-1)}
\sim \frac{2^{2m} m^{2m} e^{-2m}\sqrt{4\pi m}}
{4^m m^{2m} e^{-2m} 2\pi m\times 2m}
=\frac 1{\sqrt{\pi} m^{3/2}};\\
\sum m^{-3/2}<\infty.
$$
A: Here is an approach. I introduced a closed form for the more general case in a previous problem 

$$\sum_{i=1}^{n} i^s = \sum_{i=1}^{\infty}i^s - \sum_{i=0}^{\infty}(i+n+1)^s = \zeta(-s) - \zeta(-s, n+1)\,,$$

which gives 
$$ \sum_{i=1}^{n} i^{1/2} = \zeta(-1/2) - \zeta(-1/2, n+1)$$

$$ \implies 2\sqrt{n}-\sum_{i=1}^{n} i^{1/2} =2\sqrt{n} -\zeta(-1/2) + \zeta(-1/2, n+1) .$$

Now, to find the limit as $n$ goes to $\infty$ you need to use the identity $ 25.11.43 $.
