Compute the sum \(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n}\)

I can wrie the terms as 1/3 + 2/9 + 3/27 + ...., but I don't know what to do after that.

Guest Jul 8, 2020

#1**0 **

A fun solution:

\(\displaystyle\sum_{k = 1}^n 1 = n\)

Keeping that in mind:

\(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n} = \sum_{n=1}^\infty \sum_{k=1}^n 3^{-n}\)

By Fubini's theorem, the sums are interchangeable.

\(\displaystyle \sum_{n = 1}^\infty \frac{n}{3^n} = \sum_{k=1}^\infty \sum_{n=k}^\infty 3^{-n} = \sum_{k =1 }^\infty \dfrac{3^{-k}}{1 - \dfrac13} = \dfrac32 \cdot \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac34\)

MaxWong Jul 8, 2020

#2**0 **

An easy-to-understand solution:

After writing the terms, let S be the original sum and consider S/3.

S | = | 1/3 | + | 2/9 | + | 3/27 | + | ... |

S/3 | = | 1/9 | + | 2/27 | + | ... |

Subtracting gives:

S | = | 1/3 | + | 2/9 | + | 3/27 | + | ... |

S/3 | = | 1/9 | + | 2/27 | + | ... | ||

2S/3 | = | 1/3 | + | 1/9 | + | 1/27 | + | ... |

Now 2S/3 is the sum of geometric series 1/3 + 1/9 + 1/27 + ...

\(\dfrac{2S}3 = \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac12\\ S = \dfrac34\)

Therefore the required sum is 3/4.

MaxWong
Jul 8, 2020