Find $\lim\limits_{n\rightarrow\infty}\sqrt[n]{\sum_{k=0}^{n} \frac{(-1)^k}{k+1}\cdot2^{n-k}\cdot\binom{n }{k}}$

Question

Find

$$\lim_{n\rightarrow\infty}\sqrt[n]{\sum\limits_{k=0}^{n} \frac{(-1)^k}{k+1}\cdot2^{n-k}\cdot\binom{n }{k}}$$

My attemt: By the binomia theorem, we have $$(1-x)^n={\sum_{k=0}^{n} \binom{n }{k} (-1)^k x^k} $$ from here how can I manipulate this to get the limit?

Answer 1

We have $$ \begin{aligned} ( 2 - x) ^{ n} = \sum_{ k = 0}^{ n} \binom{ n}{ k} ( -1) ^{ k} x^{ k} 2^{ n - k} .\end{aligned} $$ Now integrating and dividing yields $$ \begin{aligned} \frac{1}{ t} \int_{ 0}^{t } ( 2 - x) ^{ n}\, dx = \frac{1}{ t} \sum_{ k = 0}^{ n} \binom{ n}{ k} ( -1) ^{ k} \frac{ t ^{ k + 1}}{ k + 1} 2^{ n - k} = \sum_{ k = 0}^{ n} \binom{ n}{ k} \frac{ ( -1) ^{ k}}{ k + 1} t ^{ k } 2^{ n - k} .\end{aligned} $$

Now $$ \begin{aligned} \int_{ 0}^{ t} ( 2 - x) ^{ n}\, dx = \frac{ -( 2 - t) ^{ n + 1} + 2^{ n + 1}}{ n + 1} .\end{aligned} $$ Inserting $ t = 1$ you obtain $$ \begin{aligned} \frac{ 2^{ n + 1}- 1 }{ n + 1} \end{aligned} $$ so taking $ n$th root you get $ 2$ as the limit.

Answer 2

\begin{align} \sum_{k=0}^{n} \frac{(-1)^k}{k+1}2^{n-k}\binom{n}{k} &= \frac{2^n}{n+1} \sum_{k=0}^{n} \left(-\frac{1}{2}\right)^k\binom{n+1}{k+1} \\ &= \frac{2^n}{n+1} \sum_{k=1}^{n+1} \left(-\frac{1}{2}\right)^{k-1}\binom{n+1}{k} \\ &= \frac{-2^{n+1}}{n+1} \sum_{k=1}^{n+1} \left(-\frac{1}{2}\right)^k\binom{n+1}{k} \\ &= \frac{-2^{n+1}}{n+1} \left(\sum_{k=0}^{n+1} \left(-\frac{1}{2}\right)^k\binom{n+1}{k} - 1\right) \\ &= \frac{-2^{n+1}}{n+1} \left(\left(1-\frac{1}{2}\right)^{n+1} - 1\right) \\ &= \frac{2^{n+1}-1}{n+1} \end{align}