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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.

 Jul 8, 2020
 #1
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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.

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

 Jul 8, 2020
 #2
avatar+8336 
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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

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