22
$\begingroup$

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!

$\endgroup$
  • 5
    $\begingroup$ I'm surprised that infinite series are addressed in eighth grade. $\endgroup$ – Michael Albanese Jul 24 '14 at 13:24
  • 5
    $\begingroup$ @MichaelAlbanese but I learned infinite geometric series $\endgroup$ – L Lawliet Jul 24 '14 at 14:03
  • 3
    $\begingroup$ Here is a related post. $\endgroup$ – David Mitra Jul 24 '14 at 14:09
  • 3
    $\begingroup$ Kids are required to do more mathematically earlier now with the common core instituted... $\endgroup$ – Lalaloopsy Jul 24 '14 at 14:12
  • 1
    $\begingroup$ @DavidMitra Thanks for your answer and link. Very helpful :) $\endgroup$ – L Lawliet Jul 24 '14 at 14:12
22
$\begingroup$

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 $$ {1\over 5^3}+{1\over 5^5}+{1\over 5^7}+\cdots\ =\ {1\over5}\Bigl( {1\over 25}+{1\over 25^2}+{1\over25^3}\cdots \Bigr) ={1\over 5}{1/25\over1-1/25}. $$

$\endgroup$
50
+50
$\begingroup$

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 $$ S-\frac{S}{5^2}=\frac{1}{5}+\color{blue}{\text{infinite geometric progression}} $$

$\endgroup$
  • 7
    $\begingroup$ This is really nice ! $\endgroup$ – Claude Leibovici Jul 24 '14 at 13:28
  • 6
    $\begingroup$ Beat me again! :( $\endgroup$ – MonK Jul 24 '14 at 13:28
  • 2
    $\begingroup$ Thanks @ClaudeLeibovici. Next time you will beat me Sid. $\ddot\smile$ $\endgroup$ – Tunk-Fey Jul 24 '14 at 13:31
26
$\begingroup$

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 $f\left(\frac15\right)$.

$\endgroup$
  • 5
    $\begingroup$ What is this Sir? I don't get it $\endgroup$ – L Lawliet Jul 24 '14 at 13:29
  • 4
    $\begingroup$ Sorry Sir, I don't understand but thanks for your answer. I appreciate it :) $\endgroup$ – L Lawliet Jul 24 '14 at 13:37
  • 4
    $\begingroup$ This solution requires calculus and analysis, which 8th graders typically aren't yet taught. $\endgroup$ – NovaDenizen Jul 24 '14 at 13:40
  • 8
    $\begingroup$ 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. $\endgroup$ – zarathustra Jul 24 '14 at 13:47
  • 1
    $\begingroup$ Really nice solution. However, shouldn't you require at least $x<1$ in order for the series to converge? $\endgroup$ – Daniel Robert-Nicoud Jul 25 '14 at 8:38
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.