Why should this be an increasing sequence? [duplicate]

The sequence $y_n = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3} \dots +\frac{1}{2n}$. Is $y_n$ convergent or divergent?

What I did:

From Mathematical Induction I derived that $y_n$ is an increasing sequence.
Now, let $y_n$ is bounded above.
Let a set Y = {$y_n$: n ∈ N}. If Y is bounded above, it has to have a supremum. Hence let $M=supY$
Therefore, $y_n \le M$ for every n ∈ N.
Hence, given $\epsilon \gt 0$, there exists some $y_k \in\Bbb Y$ such that $y_k \gt M-\epsilon$
$y_k+\epsilon \gt M$
$M \lt \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \dots +\frac{1}{2k}+\epsilon$
Let $\epsilon = \frac{1}{2(k+1)}$ Then,
$M \lt \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3} \dots +\frac{1}{2k}+\frac{1}{2(k+1)}$ $\lt$ $\frac{1}{k+2}+\frac{1}{k+3} \dots +\frac{1}{2k}+\frac{1}{2(k+1)}$=$\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2} \dots +\frac{1}{2(k+1)}$ =$y_{k+1}$
Hence $M \lt y_{k+1}$ for some $y_{k+1} \in\Bbb Y$. This is a contradiction which means, Y has no supremum which means Y is not bounded above. Hence Divergent.
I don't understand where did I go wrong in this proof because clearly Y is bounded above because $\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3} \dots +\frac{1}{2n}$ $\lt$ $\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+1} \dots +\frac{1}{n+1}$$=$$\frac{n}{n+1}$ $\lt$$1$.
Hence 1 is clearly an upper bound.So, what is wrong with that proof?

Actually I'm not good with proving and I'm trying to improve my reasoning. Help is appreciated

Solution 1:

Let $u_n=\frac{1}{n+1}+\ldots+\frac{1}{2n}$, we have $$u_n=\sum_{k=n+1}^{2n}{\frac{1}{k}}=\sum_{k=1}^n{\frac{1}{n+k}}=\frac{1}{n}\sum_{k=1}^n{\frac{1}{1+\frac{k}{n}}}\underset{n\rightarrow+\infty}{\longrightarrow}\int_0^1\frac{dx}{1+x}=\ln 2$$ What is wrong in your proof is that you wrote $\frac{1}{k+1}+\ldots+\frac{1}{2(k+1)}<\frac{1}{k+2}+\ldots\frac{1}{2(k+1)}$.

Solution 2:

$$y_{n+1}-y_n=\frac{1}{n+2}+...+\frac{1}{2n+2}-$$ $$\frac{1}{n+1} -...-\frac{1}{2n}=$$

$$\frac{1}{2n+2}+\frac{1}{2n+1}-\frac {1}{n+1}=$$

$$\frac{ 1}{(2n+1)(2n+2)}>0$$

On the other hand

$$n\frac{1}{2n}\le y_n \le n\frac{1}{n+1}\le 1$$

thus $(y_n)$ is increasing and bounded and then convergent.