How to find $\lim_{n\to\infty}\left(\frac{1}{n}+\frac{1}{2^n-1}\sum_{k=1}^{n} \frac{n \choose k}{k}\right)$ I encountered a problem when I was doing my homework where I had to find this$$\lim_{n\to\infty}\left(\frac{1}{n}+\frac{1}{2^n-1}\sum_{k=1}^{n} \frac{n \choose k}{k}\right)$$
I guess it may be about $0.1$ but till now I have no idea about it after I tried all the methods I've learned. (I'm just a freshman so I know little in this area)
 A: The limit can be split as $$\lim_{n \to \infty} \frac{1}{n} + \lim_{n \to \infty} \left(\frac{1}{2^n-1} \sum_{k=1}^n \frac{\binom{n}{k}}{k}\right)$$
The first limit is $0$, which is easy to see. Using that $\frac{1}{k} = \int_0^1 x^{k-1} dx$, the sum in the second limit is $$\sum_{k=1}^n \binom{n}{k} \int_0^1 x^{k-1} dx = \int_0^1 \sum_{k=1}^n \binom{n}{k} x^{k-1} dx = \int_0^1 \frac{(x+1)^n-1}{x} dx$$
Then the second limit is $$\lim_{n \to \infty}\frac{\int_0^1 \frac{(x+1)^n-1}{x}dx}{2^n-1} \le \lim_{n \to \infty} \frac{\int_0^1 \frac{(x+1)^n-1}{x}dx}{2^n} = \lim_{n \to \infty} \int_0^1 \frac{\left(\frac{x+1}{2}\right)^n-\frac{1}{2^n}}{x}dx$$
Since the integrand converges uniformly to $0$ for $x \in [0, 1)$ and the integral limits are finite, the integral is then $0$. Therefore, the second limit would be less than or equal to $0$. However, the limit must be nonnegative because it is a sum of nonnegative terms, so the limit must be $0$.
Therefore $$\lim_{n\to\infty}\left(\frac{1}{n}+\frac{1}{2^n-1}\sum_{k=1}^n \frac{\binom{n}{k}}{k}\right) = 0$$
Edit: As pointed out by metamorphy in the comments, this does not completely hold up since the integrand does not converge to $0$ uniformly for $x \in [0, 1]$. Instead, splitting the integral as $$\int_0^{1-\epsilon} \frac{\left(\frac{x+1}{2}\right)^n-\frac{1}{2^n}}{x} dx+\int_{1-\epsilon}^1
\frac{\left(\frac{x+1}{2}\right)^n-\frac{1}{2^n}}{x} dx$$
where $\epsilon$ approaches $0$ from the positive side. Then the integrand would converge uniformly to $0$ for $x \in [0, 1-\epsilon]$ so the first integral would converge to $0$ as well. The integrand in the second integral would converge to a finite limit, but since the lower integration bound approaches the upper bound of $1$, the second integral would approach $0$. Therefore, the overall integral would approach $0$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{{1 \over n}
+ {1 \over 2^{n} - 1}\sum_{k = 1}^{n}{{n \choose k} \over k}}} =
\lim_{n \to \infty}\bracks{%
{1 \over 2^{n} - 1}\sum_{k = 1}^{n}
{{n \choose k} \over k}}
\end{align}

Note that
\begin{align}
\sum_{k = 1}^{n}{{n \choose k} \over k} & =
\int_{0}^{1}\sum_{k = 1}^{n}{n \choose k}t^{k - 1}
\,\dd t =
\int_{0}^{1}{\pars{1 + t}^{n} - 1 \over
t}\,\dd t
\\[2mm] & =
\int_{1}^{2}{t^{n} - 1 \over  t - 1}\,\dd t =
\int_{1}^{2}\sum_{k = 0}^{n - 1}t^{k}\,\dd t =
\sum_{k = 0}^{n - 1}{2^{k + 1} - 1\over k + 1}
\\[5mm] & \mbox{and}\,\,\,
\lim_{n \to \infty}{1 \over 2^{n} - 1}
\sum_{k = 0}^{n - 1}{2^{k + 1} - 1\over k + 1}
\\[5mm] & =
\lim_{n \to \infty}
{\pars{2^{n + 1} - 1}/\pars{n + 1} \over 2^{n}} = \color{red}{\large 0}
\end{align}
\begin{align}
&\mbox{}
\\
\implies \bbx{\bbox[5px,#ffd]{\lim_{n \to \infty}\bracks{{1 \over n}
+ {1 \over 2^{n} - 1}\sum_{k = 1}^{n}{{n \choose k} \over k}}} = 0} \\ &
\end{align}
