Using Fatou's lemma to find the limit of a series. I want to find the limit of the following series $$\lim_{n\to\infty}  \sum_{k=1}^{n} \frac{1}{k}(1-\frac{1}{n})^k $$ Isn't it just the harmonic series?
So how can I prove that the harmonic series is divergent using Fatou's lemma.
 A: \begin{align*}
\liminf_{n}\sum_{k=1}^{n}\dfrac{1}{k}\left(1-\dfrac{1}{n}\right)^{k}&=\liminf_{n}\sum_{k=1}^{\infty}\chi_{1\leq k\leq n}\dfrac{1}{k}\left(1-\dfrac{1}{n}\right)^{k}\\
&\geq\sum_{k=1}^{\infty}\liminf_{n}\chi_{1\leq k\leq n}\dfrac{1}{k}\left(1-\dfrac{1}{n}\right)^{k}\\
&=\sum_{k=1}^{\infty}\dfrac{1}{k}\\
&=\infty.
\end{align*}
But $(1-1/n)$ being increasing, so
\begin{align*}
\liminf_{n}\sum_{k=1}^{n}\dfrac{1}{k}\left(1-\dfrac{1}{n}\right)^{k}=\lim_{n}\sum_{k=1}^{n}\dfrac{1}{k}\left(1-\dfrac{1}{n}\right)^{k}.
\end{align*}
A: I don't see where your idea of using this to prove divergence of the harmonic series comes from. I'd say it's the other way around.
This is $\sum\limits_{k\in\Bbb N} f_n(k)$ with $$f_n(k)=\begin{cases}0&\text{if }k=0\lor k>n\\ \frac1k\left(1-\frac1n\right)^k&\text{if }1\le k\le n\end{cases}$$
Since the functions are non-negative and $f_n(k)\to \begin{cases}0&\text{if }k=0\\\frac1k&\text{if }k\ge1\end{cases}$, by Fatou $\liminf\limits_{n\to\infty}\sum\limits_{k\in \Bbb N}f_n(k)=\infty$
