Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$

Question

What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?

Here are a few remarks:

• Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$. Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.

• Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k} &=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\ &= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\ \end{align}$$

Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed $$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\ &\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\ &\small= o\left(\frac{2^n}n \right) \end{align}$$

Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$

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This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).

Answer 1

Write the expression as

$$\frac{\sum_{k=1}^{n}2^k/k}{2^n/n}.$$

Note the denominator $\to \infty.$ That rings the Stolz-Cesaro bell, so consider

$$\frac{2^{n+1}/(n+1)}{2^{n+1}/(n+1) -2^n/n} = \frac{1}{1 -(n+1)/(2n)} \to \frac{1}{1/2} = 2.$$

By Stolz-Cesaro, the desired limit is $2.$

Answer 2

Let $S_n = \frac{n}{2^n} \sum_{k=1}^{n} \frac{2^k}{k}$. It has the following trivial lower bound.

$$ 2 - \frac{1}{2^n} = \sum_{k=1}^{n} \frac{1}{2^{n-k}} \leq S_n. $$

For an upper bound, fix $r \in (0, 1)$ and let $N = N(r,n) = \lfloor rn \rfloor$. Then for $n$ large, we have $1 < N < n$ and hence

\begin{align*} S_n &= \frac{n}{2^n} \sum_{k=1}^{N} \color{red}{\frac{2^k}{k}} + \frac{n}{2^n} \sum_{k=N+1}^{n} \color{blue}{\frac{2^k}{k}} \\ &\leq \frac{n}{2^n} \sum_{k=1}^{N} \color{red}{2^k}+ \frac{n}{2^n} \sum_{k=N+1}^{n} \color{blue}{\frac{2^k}{rn}} \\ &= \frac{n}{2^n} (2^{N+1} - 1) + \frac{1}{r 2^n}(2^{n+1} - 2^{N+1}) \\ &\leq n 2^{-(1-r)n+1} + \frac{2}{r}. \end{align*}

For the red terms, we utilized the trivial bound $\frac{1}{k} \leq 1$. For the blue terms, we utilized the fact that $k \geq N+1$ implies $k \geq nr$. Thus it follows that

$$ 2 \leq \liminf_{n\to\infty} S_n \leq \limsup_{n\to\infty} S_n \leq \frac{2}{r}. $$

Taking $r \uparrow 1$ gives the wanted conclusion.

Answer 3

$\begin{array}\\ \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} &=n\sum_{k=1}^n \frac{2^{k-n}}{k}\\ &=n\sum_{k=0}^{n-1} \frac{2^{-k}}{n-k}\\ &=\sum_{k=0}^{n-1} \frac{2^{-k}}{1-k/n}\\ \text{so}\\ \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}-2 &=\sum_{k=0}^{n-1} 2^{-k}(\frac1{1-k/n}-1)-\frac1{2^{n-1}}\\ &=\sum_{k=0}^{n-1} 2^{-k}(\frac{k/n}{1-k/n})-\frac1{2^{n-1}}\\ \end{array} $

If $k \le cn$, $2^{-k}(\frac{k/n}{1-k/n}) \le 2^{-k}\frac{c}{1-c} $ so $\sum_{k=0}^{\lfloor nc \rfloor} 2^{-k}(\frac{k/n}{1-k/n}) \le 2\frac{c}{1-c} \lt 4c$ if $0 < c < \frac12$.

If $k > cn$,

$\begin{array}\\ \sum_{k=cn}^{n-1} 2^{-k}(\frac{k/n}{1-k/n}) &<\sum_{k=cn}^{n-1} 2^{-cn}(\frac{k/n}{1-k/n})\\ &=2^{-cn}\sum_{k=cn}^{n-1} (\frac{k}{n-k})\\ &<2^{-cn}\sum_{k=cn}^{n-1} n\\ &<n^22^{-cn}\\ &<e^{-cn \ln 2 + 2\ln n}\\ \end{array} $

For any fixed $c > 0$, $-cn \ln 2 + 2\ln n \to -\infty$ as $n \to \infty$, so that $e^{-cn \ln 2 + 2\ln n} \to 0$.

By first choosing $c$ small and then $n$ large, both $4c$ and $e^{-cn \ln 2 + 2\ln n}$ can be made as small as we want, so that $\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}-2 $ can be made as small as we want, so that $\lim_{n \to \infty}\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} =2 $.

Answer 4

Let $$ S_n=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}. \tag{1}$$ Clearly $$ S_n=\sum_{k=1}^n\frac{n}{k}\frac1{2^{n-k}}\ge \sum_{k=1}^n\frac1{2^{n-k}}=2-\frac{1}{2^n}. $$ Define \begin{eqnarray} f_n(x)=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}\left(\frac12x\right)^k \end{eqnarray} and then $f_n(0)=0,f_n(2)=S_n$ and $$ f_n'(x)=\frac{n}{2^n}\sum_{k=1}^nx^{k-1}=\frac{n}{2^n}\frac{1-x^n}{1-x}.$$ So \begin{eqnarray} S_n&=&\frac{n}{2^n}\int_0^2\frac{1-x^n}{1-x}dx\\ &=&\frac{n}{2^n}\int_0^1\frac{1-x^n}{1-x}dx+\frac{n}{2^n}\int_1^2\frac{x^n-1}{x-1}dx. \end{eqnarray} Since $$ \int_0^1\frac{1-x^n}{1-x}dx=\ln n+\gamma+o(1),$$ one $$ \frac{n}{2^n}\int_0^1\frac{1-x^n}{1-x}dx=o(1). $$ Noting that if $x\in[1,2]$, then $x-1\ge1$ and $x^n-1\le x^n$, one has $$ S_n\le o(1)+\frac{n}{2^n}\int_1^2\frac{x^n-1}{x-1}dx\le o(1)+\frac{n}{2^n}\int_1^2x^ndx=o(1)+\frac{n}{2^n}\frac{1}{n+1}(2^{n+1}-1).\tag{2}$$ From (1) and (2), one has $$ 2-\frac{1}{2^n}\le S_n\le o(1)+\frac{n}{2^n}\frac{1}{n+1}(2^{n+1}-1).$$ Letting $n\to\infty$, one has $$ \lim_{n\to\infty}S_n=2. $$

Answer 5

Let $f(x)=\sum_{k=1}^n\frac{x^k}{k}$. Then, we can write

$$f(x)=\sum_{k=1}^n \int_0^x t^{k-1}\,dk=\int_0^x \frac{1-t^n}{1-t}\,dt$$

so that $f(2)=\sum_{k=1}^n\frac{2^k}{k}=\int_0^2 \frac{1-t^n}{1-t}\,dt$.

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Next, we choose $0<\delta<1$ and write

$$\begin{align} \frac{n}{2^n}\int_0^2 \frac{1-t^n}{1-t}\,dt&=\frac{n}{2^n}\int_0^{2-\delta}\frac{1-t^n}{1-t}\,dt+\frac{n}{2^n}\int_{2-\delta}^2\frac{t^n-1}{t-1}\,dt \tag 1 \end{align}$$

We can apply the mean-value theorem for the second term on the right-hand side of $(1)$ to reveal that for some $\xi_n\in [2-\delta,2]$

$$\begin{align} \frac{n}{2^n}\int_{2-\delta}^2 \frac{t^n-1}{t-1}\,dt&=\frac{n}{2^n(\xi_n-1)}\int_{2-\delta}^2(t^n-1)\,dt\\\\ &=\frac{n}{2^n(\xi-1)}\left(\frac{2^{n+1}-(2-\delta)^{n+1}}{n+1}+\delta\right) \tag 2 \end{align}$$

Letting $n\to \infty$ in $(2)$ yields

$$\lim_{n\to \infty}\frac{n}{2^n}\int_{2-\delta}^2 \frac{t^n-1}{t-1}\,dt=\frac{2}{\xi-1}$$

For any given $\epsilon>0$, we can choose $0<\delta<1$ so small that $\left|\frac{2}{\xi_n-1}-2\right|<\epsilon$. Now, we proceed with that $0<\delta<1$ fixed.

Noting that $\frac{1-t^n}{1-t}$ is positive and monotonically increasing on $[0,2]$, the first term on the right-hand side of $(1)$ is bounded above by $\frac{n}{2^n}\frac{(2-\delta)^n-1}{1-\delta}\to 0$ as $n\to \infty$.

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Putting it all together, we obtain the coveted limit

$$\lim_{n\to \infty}\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}=2$$

Answer 6

I encountered this limit when computing $$ \sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^n}\sum_{k=0}^n\frac{2^k}{k+1} $$ in equation $(8)$ of this answer, and for which an asymptotic expansion is given in equation $(6)$ of this answer: $$ \begin{align} \frac{n}{2^n}\sum_{k=1}^n\frac{2^k}k &=\sum_{k=1}^n\frac1{\binom{n-1}{k-1}}\\ &\sim2+\frac2{n-1}+\frac4{(n-1)(n-2)}+\frac{12}{(n-1)(n-2)(n-3)}\\ &+\frac{48}{(n-1)(n-2)(n-3)(n-4)}+O\left(\frac1{n^5}\right) \end{align} $$