# 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!

• I'm surprised that infinite series are addressed in eighth grade. – Michael Albanese Jul 24 '14 at 13:24
• @MichaelAlbanese but I learned infinite geometric series – L Lawliet Jul 24 '14 at 14:03
• Here is a related post. – David Mitra Jul 24 '14 at 14:09
• Kids are required to do more mathematically earlier now with the common core instituted... – Lalaloopsy Jul 24 '14 at 14:12

Informally:

You're taking the sum of the row sums of

$\ \ \ \displaystyle{1\over 5^{\phantom 1}}$

$\ \ \ \displaystyle{1\over 5^{ 3}} \ \ \ \ \displaystyle{1\over 5^{ 3}}\ \ \ \ \displaystyle{1\over 5^{ 3}}$

$\ \ \ \displaystyle{1\over 5^{ 5}} \ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \vdots$

Take the sum of the column sums instead.

Towards this end, note, for example, that $${1\over 5^3}+{1\over 5^5}+{1\over 5^7}+\cdots\ =\ {1\over5}\Bigl( {1\over 25}+{1\over 25^2}+{1\over25^3}\cdots \Bigr) ={1\over 5}{1/25\over1-1/25}.$$

Hint :

Let $$S=\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots\tag1$$ Dividing $(1)$ by $5^2$, we obtain $$\frac{S}{5^2}=\frac{1}{5^3}+\frac{3}{5^5}+\frac{5}{5^7}+\frac{7}{5^9}+\frac{9}{5^{11}}+\cdots\tag2$$ Subtracting $(2)$ from $(1)$, we obtain $$S-\frac{S}{5^2}=\frac{1}{5}+\color{blue}{\text{infinite geometric progression}}$$

• This is really nice ! – Claude Leibovici Jul 24 '14 at 13:28
• Beat me again! :( – MonK Jul 24 '14 at 13:28
• Thanks @ClaudeLeibovici. Next time you will beat me Sid. $\ddot\smile$ – Tunk-Fey Jul 24 '14 at 13:31

Let

\begin{align}f(x)&=\sum_{n=0}^\infty(2n+1)x^{2n+1}\\&=x\sum_{n=0}^\infty(2n+1)x^{2n}\\&=x\frac{d}{dx}\left(\sum_{n=0}^\infty x^{2n+1}\right)\\&=x\frac{d}{dx}\left( \frac{x}{1-x^2}\right)\\&=x\frac{x^2+1}{(1-x^2)^2}\end{align} and notice that the desired sum is $f\left(\frac15\right)$.

• What is this Sir? I don't get it – L Lawliet Jul 24 '14 at 13:29
• Sorry Sir, I don't understand but thanks for your answer. I appreciate it :) – L Lawliet Jul 24 '14 at 13:37
• This solution requires calculus and analysis, which 8th graders typically aren't yet taught. – NovaDenizen Jul 24 '14 at 13:40
• For those who dislike the use of higher math knownledge, this has been discussed on Meta: meta.math.stackexchange.com/questions/11419/…. The purpose of the answers is to help the whole community, and since an appropriate answer was already posted, @SamiBenRomdhane's maybe helpful for other users. – zarathustra Jul 24 '14 at 13:47
• Really nice solution. However, shouldn't you require at least $x<1$ in order for the series to converge? – Daniel Robert-Nicoud Jul 25 '14 at 8:38

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