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} $.
 A: Your reasoning would be correct if $y_n<\frac{1}{n}$, which is not true. The sequence actually diverges. To prove it, we use the fact that for $k\in[\![1,n-1]\!]$,
\begin{align}
    \frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n} &\geq \frac{1}{kn+n}+\frac{1}{kn+n}+...+\frac{1}{kn+n} \\
         &\geq \frac{n}{n(k+1)}=\frac{1}{k+1}
\end{align}
Then, we can rewrite $y_n$ using such partial sums:
\begin{align}
    y_n&=\sum_{k=1}^{n-1} \frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n}\\
        &\geq\sum_{k=1}^{n-1}\frac{1}{k+1}
\end{align}
This sum diverges when $n\to\infty$, thus $y_n$ also diverges.
A: 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)
$.
A: If you know that
$$\gamma_n=\frac{1}{1}+\frac{1}{2}+..+\frac{1}{n}- \ln(n) \,,$$ is convergent then your sequence is exactly 
$$\gamma_{n^2}-\gamma_n+ \ln(n) \,.$$
As $\gamma_{n^2}-\gamma_n \to 0$ and $\ln(n) \to \infty$ it follows that your sequence diverges to $\infty$. 

Edit With the change, your sequence is 
$$\gamma_{2n}-\gamma_n+\ln(2n)-\ln(n)= \gamma_{2n}-\gamma_n+\ln(2)$$
which is convergent.
A simpler solution
your sequence is also
$$\frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$$
which is the Riemann Sum associated to $f(x)=\frac{1}{x}$ on $[1,2]$ with $x_k=x_k^*=1+\frac{k}{n}$.
Both solutions also Yield $\ln(2)$ as the limit.
Third solution
$$\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}=\frac{1}{1}\frac{1}{2}+..+\frac{1}{2n-1}-\frac{1}{2n}$$
is a well known identity, pretty standard induction problem. Then you can use the Alternating Series Test.
A: Here is how to proceed using the Riemann sum idea 
$$ y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}=\sum_{i=1}^{n}\frac{1}{n+i}$$
$$=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}} \longrightarrow_{n\to \infty} \int_{0}^{1}\frac{dx}{1+x}=\dots.$$
A: The sum
$$y_n:=\sum_{k=n+1}^{k=n^2}\frac 1 k = \Psi(n^2)-\Psi(n+1),$$
where $\Psi(x):= \frac {\Gamma'(x)} {\Gamma(x)} \sim \ln  \left( x \right) -\frac 1 2\,{x}^{-1}-\frac1 {12}\,{x}^{-2}+O \left( {x}^{-4}
 \right)
$ as $x \to \infty $. In view of this $y_n \to \infty$ as $n \to \infty.$
A: 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.
