What is the result of $\sum\limits_{n=1}^{\infty}(n-1)(\frac{5}{6})^{n-1}$?

Question

In wolfram alpha it says it's 30 but I cannot figure out why.

Oh well its already answered, I think its duplicate – jeea Apr 01 '20 at 18:56 — jeea, Apr 01 '20 at 18:56

Z Ahmed · Accepted Answer · 2020-04-01T19:17:05.357

2

$$S=\sum_{n=1}^{\infty} (n-1) (5/6)^{n-1}= \sum_{m=0}^{\infty} m (5/6)^m$$ Take infinite G.P $$(1-x)^{-1}=\sum_{k=0}^{\infty} x^k,~~ |x|<1$$ D.w.r.t.$x$ and multiply by $x$ on both sides to get $$\sum_{k=0}^{\infty}k x^k=\frac{x}{(1-x)^2}$$ Using this we get $$S=\frac{5/6}{(1-5/6)^2}=30.$$

edited Apr 01 '20 at 19:17

answered Apr 01 '20 at 18:59

Z Ahmed

43,235

You forgot to square the denominator in the last line. – Magdiragdag Apr 01 '20 at 19:11
Thanks I edited it up. – Z Ahmed Apr 01 '20 at 19:17

What is the result of $\sum\limits_{n=1}^{\infty}(n-1)(\frac{5}{6})^{n-1}$?

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