Sum of $\frac{1}{5}-\frac{2}{5^2}+\frac{3}{5^3}-\frac{4}{5^4}+…$

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?

• – lab bhattacharjee Jun 19 '16 at 15:42
• @graydad Yes, Thanx for pointing that out. – Sid1234 Jun 19 '16 at 15:45
• Even-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

$$S=\frac 15-\frac2{5^2}+\frac 3{5^3}-\dots\tag 1$$ and $$\frac S5=\frac 1{5^2}-\frac2{5^3}+\frac 3{5^4}-\dots\tag 2$$

Now, $(1)+(2)$ yields $$S+\frac S5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots\\\implies \frac{6S}5=\frac 15-\frac1{5^2}+\frac1{5^3}+\dots=\frac 16\\\implies S=\frac5{36}.$$

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 $x=-\dfrac15$.