<s>[INST] 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!
 [/INST] 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 $$S=\left(\frac 15\right)\left[\frac {1+\left(\frac 15\right)^2}{(1-\left(\frac 15\right)^2)^2}\right]=\left(\frac 15\right)\left(\frac {5^2+1}{(5-\frac 15)^2}\right)=\left(\frac 15\right)\left(\frac {25*26}{576}\right)=\frac {65}{288}$$
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