Poly and logarithms

Calculus Level 4

∑ n = 1 ∞ 1 3 n , ∑ n = 1 ∞ n 3 n , ∑ n = 1 ∞ n 2 3 n , ∑ n = 1 ∞ n 3 3 n \sum_{n=1}^\infty \dfrac1{3^n} , \quad \sum_{n=1}^\infty \dfrac n{3^n} , \quad \sum_{n=1}^\infty \dfrac{n^2} {3^n} , \quad \sum_{n=1}^\infty \dfrac{n^3}{3^n}

Using method of differences, one can prove that none of the above series is an integer.

Is it also true that ∑ n = 1 ∞ n 4 3 n \displaystyle \sum_{n=1}^\infty \dfrac{n^4}{3^n} is not an integer?

Yes, it is true No, it is not true

Pi Han Goh by Pi Han Goh , 30, Malaysia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Chew-Seong Cheong
Nov 27, 2016

Let S k = ∑ n = 1 ∞ n k 3 n S_k = \displaystyle \sum_{n=1}^\infty \frac {n^k}{3^n}

S 0 = ∑ n = 1 ∞ 1 3 n = 1 3 ( 1 1 − 1 3 ) = 1 2 \begin{aligned} S_0 & = \sum_{n=1}^\infty \frac 1{3^n} = \frac 13 \left( \frac 1{1-\frac 13} \right) = \frac 12 \end{aligned}

S 1 = ∑ n = 1 ∞ n 3 n = ∑ n = 0 ∞ n 3 n = ∑ n = 1 ∞ n − 1 3 n − 1 = 3 ∑ n = 1 ∞ n 3 n − 3 ∑ n = 1 ∞ 1 3 n = 3 S 1 − 3 S 0 ⟹ S 1 = 3 2 S 0 = 3 4 \begin{aligned} S_1 & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{3^n} = \sum_{\color{#D61F06}n=0}^\infty \frac n{3^n} = \sum_{\color{#3D99F6}n=1}^\infty \frac {n-1}{3^{n-1}} = 3 \sum_{\color{#3D99F6}n=1}^\infty \frac n{3^n} - 3 \sum_{\color{#3D99F6}n=1}^\infty \frac 1{3^n} = 3S_1 - 3S_0 \\ \implies S_1 & = \frac 32 S_0 = \frac 34 \end{aligned}

Similarly,

S 2 = 3 ∑ n = 1 ∞ n 2 − 2 n + 1 3 n = 3 S 2 − 6 S 1 + 3 S 0 = 6 S 1 − 3 S 0 2 = 3 2 \begin{aligned} S_2 & = 3 \sum_{n=1}^\infty \frac {n^2-2n+1}{3^n} = 3S_2-6S_1+3S_0 \\ & = \frac {6S_1-3S_0}2 = \frac 32 \end{aligned}

S 3 = 3 ∑ n = 1 ∞ n 3 − 3 n 2 + 3 n − 1 3 n = 3 S 3 − 9 S 2 + 9 S 1 − 3 S 0 = 9 S 2 − 9 S 1 + 3 S 0 2 = 33 8 \begin{aligned} S_3 & = 3 \sum_{n=1}^\infty \frac {n^3-3n^2+3n-1}{3^n} = 3S_3-9S_2+9S_1-3S_0 \\ & = \frac {9S_2-9S_1+3S_0}2 = \frac {33}8 \end{aligned}

S 4 = 3 ∑ n = 1 ∞ n 4 − 4 n 3 + 6 n 2 − 4 n + 1 3 n = 3 S 4 − 12 S 3 + 18 S 2 − 12 S 1 + 3 S 0 = 12 S 3 − 18 S 2 + 12 S 1 − 3 S 0 2 = 15 \begin{aligned} S_4 & = 3 \sum_{n=1}^\infty \frac {n^4-4n^3+6n^2-4n+1}{3^n} = 3S_4-12S_3+18S_2-12S_1+3S_0 \\ & = \frac {12S_3-18S_2+12S_1-3S_0}2 = 15 \end{aligned}

No, it is not true \boxed{\text{No, it is not true}} . ∑ n = 1 ∞ n 4 3 n = 15 \displaystyle \sum_{n=1}^\infty \frac {n^4}{3^n} = 15 , an integer.

9 Helpful 0 Interesting 0 Brilliant 0 Confused
Aaghaz Mahajan
Nov 1, 2018

@Pi Han Goh Sir, your title seems to say that there is a hidden connection between the series and Polylogs......Is it so??? Could you please share it??

0 Helpful 0 Interesting 0 Brilliant 0 Confused

It is pretty straightforward to solve this once you realized that this is just a polylogarithm function. The working is essentially the same as Chew-Seong's solution, but slightly condensed.

Pi Han Goh - 2 years, 7 months ago

Log in to reply

Ohhh!!! I see now..... These are Polylogs with negative integer orders.......But even then, we need to solve the recurrence of Stirling Number's of the second kind, to reach the final expression........and,well.......that is also a tad bit tedious.....!!! Still, thanks Sir!!!

Aaghaz Mahajan - 2 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...