Calculate sum $\sum_{n=1}^\infty {n\over3^{n-1}}$

Question

Calculate sum $$\sum_{n=1}^\infty {n\over3^{n-1}}$$The result is $9\over 4$ but I don't know how to get that.

The idea is to take derivatives :) in your case you need the derivative of $x^n$ at the point $x=1/3$. — Stan Tendijck, Mar 09 '19 at 12:58
To start you should write $$\sum\limits_{n=1}^{\infty} n\cdot \left( \frac13 \right)^{n-1}$$ — callculus42, Mar 09 '19 at 13:13

score 1 · Accepted Answer · answered Mar 09 '19 at 12:54

1

We'll differentiate a geometric series. Suppose $|x|<1$; since $\sum_{n\ge 1}x^n=\frac{x}{1-x}$, $\sum_n nx^{n-1}=\frac{1}{(1-x)^2}$. Now take $x=\frac{1}{3}$.

answered Mar 09 '19 at 12:54

J.G.

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Calculate sum $\sum_{n=1}^\infty {n\over3^{n-1}}$

1 Answers1