How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$. 
I need to prove that for $n \ge 1$: 
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$$
I guess one could say that:
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{n}{n+1}
$$
However, I'm not sure this is rigorous enough (for example, in $\sum_{i=1}^n\frac{1}{n+i}$ how do we really know that the index goes from $1$ to $n$) and I think this needs to be proven via induction. 
So the base case is:
$$a_1 = \frac{1}{2} \le \frac{1}{2} = \frac{n}{n+1}$$
The step: suppose $a_n \le \frac{n}{n+1}$ then let's prove that $$a_{n+1} \le \frac{n+1}{n+2}$$
The above can be developed as:
$$
\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)} \le \frac{n}{n+1}+\frac{1}{n+2}
$$
This is where I get stuck. If I could somehow prove that the number of terms to the left $\le$ the terms to the left I would be golden. Or maybe there's another way.
 A: $$\frac{1}{n+1} \leq \frac{1}{n+1}$$
$$\frac{1}{n+2}< \frac{1}{n+1}$$
$$\cdots$$
$$\frac{1}{2n} < \frac{1}{n+1}$$ 
Now add up!
Equality comes when $n=1$.
A: HINT:
$$\dfrac1{n+1}>\dfrac1{n+r}$$  for $2\le r\le n$
A: Your inequality can be greatly improved. The sequence given by $a_n = H_{2n}-H_{n}$ is increasing (it is enough to compute $a_{n+1}-a_n$) hence $H_{2n}-H_n\leq \lim_{n\to +\infty}\left(H_{2n}-H_n\right) = \color{red}{\log 2}$.
A: This is wrong:
$$a_{n+1}=\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)}$$
You should have:
$$a_{n+1}=\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2(n+1)}$$
Note that this has $1$ extra term than $a_n$ because the series is defined using $n$ twice - once for $n+1$ and again at $2n$.
From here subtract $a_n$ and rearrange to get:
$$a_{n+1}=a_{n}-\dfrac1{n+1}
+\dfrac1{2n+1}+\dfrac1{2n+2}$$
We have:
$$\dfrac1{n+1}=\dfrac2{2(n+1)}=\dfrac1{2n+2}+\dfrac1{2n+2}$$
So the extra fractions become:
$$
\begin{align}
&-\dfrac1{n+1}+\dfrac1{2n+1}+\dfrac1{2n+2}\\
&=-\dfrac1{2n+2}-\dfrac1{2n+2}+\dfrac1{2n+1}+\dfrac1{2n+2}\\
&=-\dfrac1{2n+2}+\dfrac1{2n+1}\\
&=\dfrac1{2n+1}-\dfrac1{2n+2}\\
&=\dfrac1{(2n+1)(2n+2)}
\end{align}
$$
As $a_n \le \dfrac{n}{n+1}$ we need to prove:
$$\dfrac{n}{n+1}+\dfrac1{(2n+1)(2n+2)}\le\dfrac{n+1}{n+2}$$
Or equivalently that:
$$\dfrac1{(2n+1)(2n+2)}\le\dfrac{n+1}{n+2}-\dfrac{n}{n+1}$$
The right hand side is:
$$\dfrac{n+1}{n+2}-\dfrac{n}{n+1}=\dfrac{n^2+2n+1-n^2-2n}{(n+1)(n+2)}=\dfrac1{(n+1)(n+2)}$$
and we are done ($(n+1)(n+2)\le(2n+1)(2n+2)$ for $n\in\mathbb{N}$).
A: One question you seem to have is:
How do we know that  $$ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $$ means 
$$
\sum_{i=1}^n\frac{1}{n+i}?$$
The answer is that we don't for sure though experience suggests it does. 
So if that IS the intended meaning then
here is no ambiguity to 
$$
\sum_{i=1}^n\frac{1}{n+i} \le  \sum_{i=1}^n\frac{1}{n+1}=\frac{n}{n+1}.$$
Since that is what was requested, it seems that the assumed definition is correct.
But maybe something like $1/2+1/3+...+1/17$ is the sum of $1/p$ for $p$ prime in some range?
Here the final term is even and whatever the rule is should depend only on $n.$
If we think that the first term is intended to be the largest (which is reasonable but not explicit ) then at least it is valid that
$$ \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad  \le \frac{1}{n+1}+\frac{1}{n+1}+...+\frac{1}{n+1}.$$
