<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 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.
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