Bounding sum of reciprocals of the square roots of the first N positive integers I am trying to derive the following inequality:
$$2\sqrt{N}-1<1+\sum_{k=1}^{N}\frac{1}{\sqrt{k}}<2\sqrt{N}+1,\; N>1.$$
I understand for $N\rightarrow \infty$  the summation term diverges (being a p-series with p=1/2), which is consistent with the lower bound in this inequality being an unbounded function in $N$.  With respect to deriving this inequality, it is perhaps easier to rewrite it as
$$2\sqrt{N}-2<\sum_{k=1}^{N}\frac{1}{\sqrt{k}}<2\sqrt{N},\; N>1.$$
In this expression, the fact that the upper and lower bounds are so concisely expressed made me wonder whether there is a more basic inequality I can begin with of the form
$$a_{k}<\frac{1}{\sqrt{k}}<b_{k},\; \forall k>1,$$
from which summing through would yield the desired inequality.  That is $a_k \text{ and } b_k$ should satisfy $\sum_{k=1}^{N}a_k=2\sqrt{N}-2$ and $\sum_{k=1}^{N}b_k=2\sqrt{N}$.  However, I am stuck on how to find the appropriate $a_{k}$ and $b_{k}$ to obtain the desired bounds.  Alternatively, what other methods are at my disposal?
 A: I will give you a proof of a stronger inequality through creative telescoping. Let we start by noticing that
$$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+1/2}+\sqrt{n-1/2}}\geq\frac{1}{2\sqrt{n}}$$
by the concavity of the function $f(x)=\sqrt{x}$ over $\mathbb{R}^+$. It follows that:
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}}\color{red}{\leq} 2\sum_{n=1}^{N}\left(\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}\right) = \color{red}{\sqrt{4N+2}-\sqrt{2}}.$$
On the other hand, $ 2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}}-\frac{1}{\sqrt{n}}=\Theta\left(\frac{1}{n^{5/2}}\right)$, hence
$$ \sum_{n=1}^{N}\frac{1}{\sqrt{n}} \color{red}{\geq \sqrt{4N+2}-\sqrt{2}-C}$$
where:
$$ 0\leq C = \sum_{n\geq 1}\left(2\sqrt{n+\frac{1}{2}}-2\sqrt{n-\frac{1}{2}}-\frac{1}{\sqrt{n}}\right)=\color{green}{-\left[\sqrt{2}+\zeta\left(\frac{1}{2}\right)\right]}\leq \frac{1}{21}. $$
A: To get the lower bound, you can convert the sum to an integral. $\frac 1{\sqrt k} \gt \int_k^{k+1}\frac 1{\sqrt x}dx=2\sqrt x|_k^{k+1}=2(\sqrt {k+1}-\sqrt k)$, so $\sum_{k=1}^N \frac 1{\sqrt k} \gt \int_1^N \frac 1{\sqrt x}dx=2\sqrt N -2$  Then to get the upper limit you have the same approach.  $\frac 1{\sqrt k} \lt \int_{k-1}^{k}\frac 1{\sqrt x}dx=2\sqrt x|_{k-1}^{k}=2(\sqrt {k}-\sqrt {k-1})$ so $\sum_{k=1}^N \frac 1{\sqrt k} \lt \int_0^{N-1} \frac 1{\sqrt x}dx=\int_0^{1} \frac 1{\sqrt x}dx+\int_1^{N-1} \frac 1{\sqrt x}dx=2+2\sqrt N -2=2\sqrt N$
A: Use induction on $N$, and compose inequalities. For instance, on the lower bound, you might get that 
$$y<\sum$$
where $\sum$ is the sum by the inductive hypothesis, and you want to prove that $$x < \sum$$ so you prove that $x<y$. 
