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?

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 $x<y$. 
