Monotonicity and convergence of the sequence $a_n=\sum_{k=1}^{n}\frac{1}{k+n}$ Let we have the following sequence $(a_n)$ such that $$a_n=\sum_{k=1}^n\frac{1}{n+k}$$ How can I prove that $(a_n)$ is increasing bounded sequence, then prove it is convergent and find its limit?
 A: $$a_n = \sum_{k=1}^{n}\frac{1}{n+k} = H_{2n}-H_n $$
gives an increasing sequence since:
$$ a_{n+1}-a_n = \sum_{m=n+2}^{2n+2}\frac{1}{m}-\sum_{m=n+1}^{2n}\frac{1}{m} = \frac{1}{2n+1}-\frac{1}{2n+2}=\frac{1}{(2n+1)(2n+2)}\geq 0$$
and by summing both sides of the previous line for $n=0,1,2,\ldots$ we get:
$$ \lim_{n\to +\infty} a_n = \sum_{n\geq 0}\frac{1}{(2n+1)(2n+2)} = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}=\int_{0}^{1}\frac{dx}{1+x}=\color{red}{\log 2}.$$

Alternative: if we consider that
$$ a_n = \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}} $$
is a Riemann sum, then the convergence to $\int_{0}^{1}\frac{dx}{1+x}=\log 2$ is trivial, and the monotonicity follows from Karamata's inequality, since $\frac{1}{1+x}$ is a convex function over $[0,1]$.
A: You have $$a_n=\frac1n\sum_{k=1}^n\frac1{1+\frac kn},$$which is a Riemann sum for $\int_0^1\frac1{1+t}\,dt.$ This shows that $a_n\to\log(2)$.
Dunno about the monotonicity.
A: or use  $H_{n}=\ln{n}+\gamma+o(n)$
so
$$\lim_{n\to\infty}\sum_{k=1}^{n}\dfrac{1}{n+k}=H_{2n}-H_{n}=\ln{2}$$
A: This is a very special case
of this question of mine:
If $s_{k,m}(n) =\sum_{i=n+1}^{kn+m} \frac1{i} $ show that for $k \ge 2m+1$, $s_{k,m}(n+1)>s_{k,m}(n)$ and $s_{k,m}(n+1)-s_{k,m}(n) <\frac1{n(n+1)} $
The result I prove there is this:
Let
$s_{k,m}(n)
=\sum\limits_{i=n+1}^{kn+m}
 \frac1{i}
$.
Show that,
for $k \ge 2m+1$,
$s_{k,m}(n+1)>s_{k,m}(n)$
and
$s_{k,m}(n+1)-s_{k,m}(n)
<\frac1{n(n+1)}
$
so that
$s_{k,m}(n) < s_{k,m}(1)+1$.
This answers the OP's question
by setting
$k=2$ and $m=0$,
aside from the part about
naming the result.
