Double limit problem

$\begin{aligned} a_n & = \frac 1{2^1} + \frac 2{2^2} + \frac 3{2^3} + \frac 4{2^4} + \cdots + \frac n{2^n} \\ 2 a_n & = 1 + \frac 22 + \frac 3{2^2} + \frac 4{2^3} + \cdots + \frac n{2^{n-1}} \\ (2a_n - a_n) & = 1 + \frac 12 + \frac 1{2^2} + \frac 1{2^3} + \cdots + \frac 1{2^{n-1}} - \frac n{2^n} \\ \implies a_n & = \frac {1-\frac 1{2^n}}{1-\frac 12} - \frac n{2^n} \\ & = 2 - \frac 1{2^{n-1}} - \frac n{2^n} \end{aligned}$

Then, we have:

$\begin{aligned} \lim_{n \to \infty} \left(\frac {2^n(2-a_n)}{n+1}\right)^{n+1} & = \lim_{n \to \infty} \left(\frac {2^n\left(2-2+2^{-n+1} + n2^{-n}\right)}{n+1}\right)^{n+1} \\ & = \lim_{n \to \infty} \left(\frac {n+2}{n+1} \right)^{n+1} \\ & = \lim_{n \to \infty} \left(1+\frac 1{n+1} \right)^{n+1} \\ & = e \end{aligned}$

$\begin{aligned} \lim_{n \to \infty} a_n & = \lim_{n \to \infty} \left(2 - \frac 1{2^{n-1}} - \color{#3D99F6} \frac n{2^n}\right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = 2 - 0 - \lim_{n \to \infty} \color{#3D99F6} \frac 1{2^n\ln 2} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }n. \\ & = 2 - \color{#3D99F6} 0 \end{aligned}$

Therefore the answer is $e$ and $2$ .

Solution:

$a_n = \sum_{k = 1}^{n} \frac{k}{2^k} = \sum_{k = 1}^{n}\sum_{i = k}^{n} \frac{1}{2^i} = \sum_{k = 1}^{n}\sum_{i = k}^{n} \left( \frac{1}{2^{i - 1}} - \frac{1}{2^i} \right) = \sum_{k = 1}^{n} \left( \frac{1}{2^{k - 1}} - \frac{1}{2^n} \right) = \sum_{k = 1}^{n} \frac{1}{2^{k - 1}} - \frac{n}{2^n}$

$= 2\sum_{k = 1}^{n} \frac{1}{2^{k}} - \frac{n}{2^n} = 2\left( -\left(\frac{1}{2}\right)^n + 1\right) - \frac{n}{2^n} = 2 - \frac{n + 2}{2^n}$ . So $a_n = 2 - \frac{n + 2}{2^n}$ .

the limit of $a_n$ is given by:

$\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \left(2 - \frac{n + 2}{2^n}\right) = 2 - \lim_{n \rightarrow \infty} \left(\frac{n + 2}{2^n}\right) = 2 - \lim_{n \rightarrow \infty} \left(\frac{1}{2^n ln 2}\right) = 2$ . follow that $\lim_{n \rightarrow \infty} a_n = 2$ .

$\lim_{n \rightarrow \infty} \left(\frac{2^n(2 - a_n)}{n + 1}\right)^{(n + 1)} = \lim_{n \rightarrow \infty} \left(\frac{2^n\left(2 - \left(2 - \frac{n + 2}{2^n}\right)\right)}{n + 1}\right)^{(n + 1)} = \lim_{n \rightarrow \infty} \left(\frac{n + 2}{n + 1}\right)^{(n + 1)} = \lim_{n \rightarrow \infty} \left(1 + \frac{1}{n + 1}\right)^{(n + 1)} = e$