Convergence of series n/2^(n-1)

Question

I have to find the value of 3.9^1/2.27^1/4.81^1/8....upto infinity The series can be written as 3^£n/2^(n-1) where n is upto infinity. I solved the question and found that n/2^(n-1)=0 Is this correct? or I am doing wrong?

See https://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series — lab bhattacharjee, Feb 18 '17 at 10:39
Related : http://math.stackexchange.com/questions/333192/solve-sum-nxn and http://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn — lab bhattacharjee, Feb 18 '17 at 10:40

score 0 · Answer 1 · answered Feb 18 '17 at 10:34

0

hint: There are several ways to tackle this question, one of which is to study the series $\sum nx^n$, and this series (sum) can be found by studying again the series $\sum x^n, |x| < 1$ which is "geometric"...

answered Feb 18 '17 at 10:34

DeepSea

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Convergence of series n/2^(n-1)

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