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evaluate the sum: \(\[\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k} + \cdots \]\)

 Nov 14, 2017
 #1
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1/3 + 2/3^2 + 3/3^3 + 4/3^4 + 5/3^5 +..........+k/3^k

 

sum_(n=1)^k 3^(-n) n = 1/4 3^(-k) (-2 k + 3^(k + 1) - 3) =converges to 3/4. See here:

https://www.wolframalpha.com/input/?i=1%2F3+%2B+2%2F3%5E2+%2B+3%2F3%5E3+%2B+4%2F3%5E4+%2B+5%2F3%5E5+%2B..........%2Bk%2F3%5Ek

 Nov 14, 2017
edited by Guest  Nov 14, 2017

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