Proof that $\sum_{i=1}^n \frac{1}{n+i}$ is bounded $$\sum_{i=1}^n \frac{1}{n+i}=\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n-1)}+\frac{1}{2n}$$
I'm having problems proving that this sum is bounded.
I know it's bounded from below because it's a sum of positive numbers but I'm having trouble proving it has an upper bound. I tried finding an upper bound for every part of the sum but I always end up with something depending on $n$ which diverges.
 A: HINT
We have that $\forall n$
$$n\cdot \underbrace{\frac{1}{2n}}_{smaller}\le \overbrace{\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{n+(n-1)}+\frac{1}{2n}}^{n\,terms}\le n\cdot\underbrace{\frac{1}{n+1}}_{bigger}$$
A: Or let $$s=\sum_{i=1}^n \cfrac 1{n+i}$$
we have
\begin{align}
s^2 &<\left(\sum_{i=1}^n 1^2\middle)\middle(\sum_{i=1}^n \cfrac 1{(n+i)^2}\right)\hspace{2cm}\text{(Cauchy-Schwartz Inequality)}\\
&=n\sum_{i=1}^n \cfrac 1{(n+i)^2}\\
&<n\sum_{i=1}^n \cfrac 1{n(n+i)}\\
&=n\left(\cfrac 1n-\cfrac 1{n+1}+\cfrac 1{n+1}-\cfrac 1{n+2}+\cdots+\cfrac 1{2n-1}-\cfrac 1{2n}\right) \hspace{2mm}\text{(telescope series)}\\
&=n(\cfrac 1n-\cfrac 1{2n})=\cfrac 12
\end{align}
Hence $s< \cfrac {\sqrt{2}} 2$ and has to be bounded.
A: To prove that this series is bounded by an upper limit, would it suffice to prove that it is less than some other finite sum?
For every term, $1/(n+i) <1/(n)$. 
On summing up to $n$ terms ,
the given sum is $<n*(1/(n))$, i.e.
it is $<1$.
So it is bounded.
Please correct me if I am wrong.
