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