2

I have a finite sum: $$\sum_{k=1}^n\frac{2^{k}}{k}$$

I am wondering how to write it in closed form.

I tried this, where I will plug in $x=2$ at the end:

$$f(x):=\sum_{k=1}^nx^{k-1}=\sum_{k=0}^{n-1}x^k=\frac{1-x^{n-1}}{1-x}$$

Integrate:

$$\int f(x)\,dx=\sum_{k=1}^n\int x^{k-1}dx=\sum_{k=1}^n\frac{x^k}{k}$$

but when I try to integrate $f(x)$, I am not able to reach a closed form, i.e. I get a long and convoluted sum...

Thank you!

IV_
  • 6,964
Toffe1369
  • 65
  • 4
  • Your value for the sum which appears after "f(x):=" is off. For $n=1$ the sum should be $1$ but the expression you give for $n=1$ gives $0.$ I think the exponent in the numerator should be $n$ rather than your $n-1.$ – coffeemath May 27 '22 at 19:17
  • @coffeemath it's like that on purpose so that at end of the integration i get a summation starting at k=1 with x^k. – Toffe1369 May 27 '22 at 19:20
  • Look at the three expressions you have put after $f(x):=$ and make $n=1$ in each of the three expressions. They are not the same. – coffeemath May 27 '22 at 19:23
  • 3
    I don't think there is a closed form (at least without using special functions) for this sum, but you estimate the behaviour for large $n$ like in this post https://math.stackexchange.com/q/888354/399263 – zwim May 27 '22 at 19:23
  • @zwim ok, thank you for your help – Toffe1369 May 28 '22 at 06:07

2 Answers2

1

This sum can be represented in terms of Lerch zeta function:

$$\sum_{k=1}^n\frac{2^k}{k}=-\pi i-2^n\ \text{LerchPhi}(2,1,n)+\frac{2^n}{n}$$

IV_
  • 6,964
1

It has no simple closed form, but its asymptotic evaluation is pretty straightfoward:

$$ S(n)=\sum_{k=1}^{n}\frac{2^k}{k} = \frac{2^n}{n}+2^n\sum_{k=1}^{n-1}\frac{1}{k 2^{n-k}}=\frac{2^n}{n}+2^n[x^n]\left(\sum_{a\geq 1}\frac{x^a}{a}\sum_{b\geq 1}\frac{x^b}{2^b}\right)$$ hence

$$ S(n)=\frac{2^n}{n}+2^n [x^n]\left(-\log(1-x)\frac{x}{2-x}\right)=\frac{2^n}{n}+\frac{2^n}{2\pi i}\oint_{|x|=1-\varepsilon}\frac{-\log(1-x)}{(2-x)x^n}\,dx $$

or

$$ S(n)=\frac{2^n}{n}+2^n\operatorname*{Res}_{x=0}\frac{-\log(1-x)}{(2-x)x^{n+1}}. $$ Even without such representation, by Cesaro-Stolz

$$ \lim_{n\to +\infty}\frac{S(n)}{2^n/n} = \lim_{n\to +\infty}\frac{S(n+1)-S(n)}{2^{n+1}/(n+1)-2^n/n}=\lim_{n\to +\infty}\frac{2^{n+1}/(n+1)}{2^{(n+1)}/(n+1)-2^n/n}=2 $$ and the other terms of the asymptotic expansion can be reconstructed in a similar fashion.

Jack D'Aurizio
  • 353,855