<s>[INST] Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was 
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. That is $\frac{-1}{n} \leq \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \leq \frac{1}{n}$ Since we also proved earlier that $lim\frac{1}{n}=0$ and $lim\frac{-1}{n}= -lim\frac{-1}{n}=0$ It follows by the squeeze theorem that $lim(y_n)=0$. Would this be correct?
Edit: This what I have now.
Well this what I get so far that $ \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since  $ \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $ \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $.
 [/INST] If
$y_n=\sum\limits_{k=n+1}^{2n}\frac{1}{k}
$,
then
$\begin{array}\\
y_{n}-y_{n-1}
&=\sum\limits_{k=n+1}^{2n}\frac{1}{k}
-\sum\limits_{k=n}^{2n-2}\frac{1}{k}\\
&=\frac1{2n}+\frac1{2n-1}-\frac1{n}\\
&=\frac1{2n}+\frac1{2n-1}-(\frac1{2n}+\frac1{2n})\\
&=\frac1{2n-1}-\frac1{2n}\\
&<\frac1{2n-2}-\frac1{2n}\\
&=\frac12(\frac1{n-1}-\frac1{n})\\
\end{array}
$
and
$y_{n}-y_{n-1}
> 0$.
Summing from
$n+1$ to $m$,
$y_m-y_n
=\sum\limits_{k=n+1}^m (y_{k}-y_{k-1})
> 0
$
and
$y_m-y_n
=\sum\limits_{k=n+1}^m (y_{k}-y_{k-1})
<\sum\limits_{k=n+1}^m \frac12(\frac1{k-1}-\frac1{k})
= \frac12(\frac1{n}-\frac1{m})
< \frac1{2n}
$.
By the Cauchy criteria,
$y_n$ converges,
without showing what the limit is.
</s>