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} $.

Note that
$\frac1{k+1} 
\le \int_k^{k+1} \frac{dt}{t} 
\le \frac1{k}
$.
Summing from $n+1$ to $n^2$,
$\sum_{k=n}^{n^2-1}\frac1{k+1} 
\le \sum_{k=n}^{n^2-1}\int_k^{k+1} \frac{dt}{t} 
\le \sum_{k=n}^{n^2-1}\frac1{k}
$,
or
$\sum_{k=n+1}^{n^2}\frac1{k} 
\le \int_n^{n^2} \frac{dt}{t} 
\le \sum_{k=n}^{n^2-1}\frac1{k}
$.
Using the right-hand inequality,
$
\sum_{k=n+1}^{n^2}\frac1{k}
=-\frac1{n}+\frac1{n^2}+\sum_{k=n}^{n^2-1}\frac1{k}
\ge -\frac1{n}+\frac1{n^2}+\int_n^{n^2} \frac{dt}{t} 
>-\frac1{n}+ \ln(n^2)-\ln(n)
= \ln(n)-\frac1{n}
$
which diverges as $n \to \infty$.
If we use the left-hand inequality,
$
\sum_{k=n+1}^{n^2}\frac1{k}
\le \ln(n)
$.
