Does $\sum_{k=1}^n\frac{n}{n+k}$ converges? I want to know wether the sequence 
$$\sum_{k=1}^n\frac{n}{n+k}$$
converges. I think it does not since i can write it as
$$\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}$$
which is if i'm not mistaken $n\int_0^1 \frac{1}{1+x}dx$, hence $nln(2)$ when passing $n$ to infinity. Is it ok ?
 A: Easier answer is to use:
$$\frac{n}{n+k}\geq \frac{n}{2n}=\frac{1}{2}$$
for $0\leq k\leq n$.
So: $$\sum_{k=1}^n \frac{n}{n+k} \geq \frac{n}{2}.$$
A: It does not. By a Riemann sum argument:
$$ \lim_{n\to +\infty}\sum_{k=1}^{n}\frac{1}{k+n}=\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}=\int_{0}^{1}\frac{dx}{x+1}=\log 2,\tag{1}$$
and by the convexity of the function $\frac{1}{1+x}$ over $\mathbb{R}^+$, the Hermite-Hadamard inequality gives:
$$ \sum_{k=1}^{n}\frac{1}{k+n}=\log 2+O\left(\frac{1}{n}\right),\tag{2}$$
hence $\sum_{k=1}^{n}\frac{n}{n+k}$ behaves like $n\log(2)+O(1)$ and it is trivially divergent, as claimed.
A: Alternatively, let $$\begin{align}s_n&=\sum_{k=1}^n\dfrac{n}{n+k}\\
& =  \sum_{k=1}^n \left(1-\dfrac{k}{n+k}\right)\\
&=n-\displaystyle \sum_{k=1}^n \dfrac{k}{n+k}\\
&=n-n\sum_{k=1}^n \dfrac{1}{n}\cdot \dfrac{\dfrac{k}{n}}{1+\dfrac{k}{n}}\\& =n\left(1-\dfrac{1}{n}\displaystyle \sum_{k=1}^n\dfrac{\dfrac{k}{n}}{1+\dfrac{k}{n}}\right)\\& = n\left(1-t_n\right)\\& = r_n\cdot b_n\end{align}$$.
Now since $t_n = \dfrac{1}{n}\displaystyle \sum_{k=1}^n \dfrac{\dfrac{k}{n}}{1+\dfrac{k}{n}}\to \displaystyle \int_{0}^1 \dfrac{x}{1+x}dx=1-\ln 2$, we have:
$$\begin{align} r_n&= n \to \infty, b_n = 1-t_n \to 1-(1-\ln 2) = \ln 2 \\& \Rightarrow s_n = r_n\cdot b_n \to \infty\end{align}$$ as claimed.
