Find the sum of following series:
$S=\frac{1}{5}-\frac{2}{5^2}+\frac{3}{5^3}-\frac{4}{5^4}+....$ upto infinite terms
Could someone give me slight hint to solve this question?
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See http://math.stackexchange.com/questions/647587/sum-of-a-power-series-n-xn – lab bhattacharjee Jun 19 '16 at 15:42
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@graydad Yes, Thanx for pointing that out. – Sid1234 Jun 19 '16 at 15:45
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2Even-powered terms are not in G.P. the next even term is $-6/5^6$, not $-8/5^6$. – Thomas Andrews Jun 19 '16 at 15:47
2 Answers
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$$S=\frac 15-\frac2{5^2}+\frac 3{5^3}-\dots\tag 1$$ and $$\frac S5=\frac 1{5^2}-\frac2{5^3}+\frac 3{5^4}-\dots\tag 2$$
Now, $(1)+(2)$ yields $$S+\frac S5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots\\\implies \frac{6S}5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots=\frac 16\\\implies S=\frac5{36}.$$
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Hint. One may recall (see this) that $$ \sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2},\quad |x|<1. $$ Then apply it with $x=-\dfrac15$.
Olivier Oloa
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