Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:

Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$

This one is very hard for me. It seems unsolvable. How to calculate the series without using Wolfram Alpha? Please help me. Grazie!

This sum can be represented in the form $$S=\sum_{k=0}^\infty (2k+1)x^{2k+1}$$$$=>S=x\sum_{k=0}^\infty (2k+1)x^{2k}$$ where $x=\frac 15$. Now we look at the following geometric progression $S(x)=\sum_{k=0}^\infty x^{2k+1}$ where $|x|\lt 1$.hence $S(x)=\frac a{1-r}$ where a=x and $r=x^2$. Therefore $S(x)=\frac x{1-x^2}$. Therefore $$\sum_{k=0}^\infty x^{2k+1}=\frac x{1-x^2}$$$$=>\sum_{k=0}^\infty \frac d{dx}(x^{2k+1})=\frac d{dx} \left(\frac x{1-x^2}\right)$$$$=>\sum_{k=0}^\infty (2k+1)x^{2k}=\left[\frac {1+x^2}{(1-x^2)^2}\right]$$$$=>\sum_{k=0}^\infty (2k+1)x^{2k+1}=x\left[\frac {1+x^2}{(1-x^2)^2}\right]$$
Now in the question $x=\frac 15$. Therefore $$S=\left(\frac 15\right)\left[\frac {1+\left(\frac 15\right)^2}{(1-\left(\frac 15\right)^2)^2}\right]=\left(\frac 15\right)\left(\frac {5^2+1}{(5-\frac 15)^2}\right)=\left(\frac 15\right)\left(\frac {25*26}{576}\right)=\frac {65}{288}$$
