Prove the series $\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6$

Question

Prove the series $$\sum_{n = 1}^{ \infty} \frac 1 6 n (\frac 5 6)^{n-1} = 6.$$

I've tried various methods for proving the series:

The series is not geometric, but I see that $\frac 1 6 n (\frac 5 6)^{n-1} \rightarrow 0$. Also the series is not telescoping, as far I'm concerned.

What method could solve this ?

Hint: $f(x)=\sum x^n=1/(1-x)$, $f'(x)=\cdots$ – Martín-Blas Pérez Pinilla Apr 15 '14 at 14:26 — Martín-Blas Pérez Pinilla, Apr 15 '14 at 14:26
This post should give you some ideas. – David Mitra Apr 15 '14 at 14:30 — David Mitra, Apr 15 '14 at 14:30

Yiyuan Lee · Answer 1 · 2014-04-15T14:36:35.063

4

Here's a calculus approach:

We know that:

$$\frac{1}{1 - x} = 1 + x + x^2 + \dots$$

Differentiating both sides,

$$\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \dots$$

Now, substitute $x = \frac{5}{6}$. Then,

$$\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{\left(1 - \frac{5}{6}\right)^2}$$

So we have: $$\frac{1}{6}\sum_{n = 1}^\infty n\left(\frac{5}{6}\right)^{n-1} = \frac{1}{6\left(1 - \frac{5}{6}\right)^2}\\ = 6$$

edited Apr 15 '14 at 14:36

answered Apr 15 '14 at 14:26

Yiyuan Lee

This might be a "stupid" question, but I've only taken derivatives of finite expressions until now - is this also legal for infinite expressions ? – Andreas Lykke Iversen Apr 15 '14 at 14:31
Yes, I believe it is legal. – Yiyuan Lee Apr 15 '14 at 14:32
In what subject of mathematics where you introduced to this ? – Andreas Lykke Iversen Apr 15 '14 at 14:34
It was directly after we were taught sequences, series and calculus in high school. One can integrate the first result to deduce similarly interesting results as well. – Yiyuan Lee Apr 15 '14 at 14:35

score 1 · Answer 2 · answered Apr 15 '14 at 14:32

1

Hint:

$$6S=1+2.\frac56+3.(\frac56)^2+4.(\frac56)^3...$$

$$\frac56\times6S=0+\frac56+2.(\frac56)^2+3.(\frac56)^3...$$

Zero is just to show order of subtraction

$$S=1+\frac56+(\frac56)^2+(\frac56)^3...$$

Hope you can carry on from here.

answered Apr 15 '14 at 14:32

evil999man

Ahh well the series is telescoping into a geometric series. Thanks ! – Andreas Lykke Iversen Apr 15 '14 at 14:35
@AndreasLykkeIversen This is an AGP. the respective terms are products of an AP,GP. Their sum till n terms can be also computed from this strategy – evil999man Apr 15 '14 at 14:39

2 Answers2