Summing Infinite Series: A Shortcut Using Differentiation

[AGNuke]

Given S, an Infinite Series Summation, find \frac{1728}{485}S
S=1^2+\frac{3^2}{5^2}+\frac{5^2}{5^4}+\frac{7^2}{5^6}+...
I found out the formula for (r+1)^th term of the series, hence making the series asS=1+\sum_{r=1}^{\infty}\frac{(2r+1)^2}{(5^r)^2}
Now I have a hard time guessing what to do from now on. I expanded the numerator in summation series, 4r² + 4r + 1. This formed the GP and AGP series (from 1 and 4r respectively). Now all that is left is to find the summation of 4r²/5^2r.

By the way, I entered the series up at Wolfram|Alpha and the answer it showed is 5, which is correct.

 

[mfb]

There might be a direct way, but here is an indirect way:

$$\sum_{r=1}^\infty \frac{r^2}{5^{2r}}\\
= \frac{1}{25} + \sum_{r=2}^\infty \frac{r^2}{5^{2r}}\\
= \frac{1}{25} + \sum_{r=1}^\infty \frac{(r+1)^2}{5^{2(r+1)}}\\
= \frac{1}{25} + \frac{1}{25}\sum_{r=1}^\infty \frac{r^2+2r+1}{5^{2r}}\\
= \frac{1}{25} \left(1+\sum_{r=1}^\infty \frac{2r+1}{5^{2r}}\right) + \frac{1}{25}\sum_{r=1}^\infty \frac{r^2}{5^{2r}}$$

 

[Ray Vickson]

Given S, an Infinite Series Summation, find \frac{1728}{485}S
S=1^2+\frac{3^2}{5^2}+\frac{5^2}{5^4}+\frac{7^2}{5^6}+...
I found out the formula for (r+1)^th term of the series, hence making the series asS=1+\sum_{r=1}^{\infty}\frac{(2r+1)^2}{(5^r)^2}
Now I have a hard time guessing what to do from now on. I expanded the numerator in summation series, 4r² + 4r + 1. This formed the GP and AGP series (from 1 and 4r respectively). Now all that is left is to find the summation of 4r²/5^2r.

By the way, I entered the series up at Wolfram|Alpha and the answer it showed is 5, which is correct.

If you do know calculus, that summation would be a standard homework exercise. If
S(x) = \sum_{n=1}^{\infty} x^n,
then
x\frac{d}{dx} S(x) = \sum_{n=1}^{\infty} n x^n \\<br /> x \frac{d}{dx} \left[ x\frac{d}{dx} S(x) \right] = \sum_{n=1}^{\infty} n^2 x^n
You can find $S(x)$ in closed form, so you can do all the calculations explicitly. Then, of course, you set x = 1/5.

 

[AGNuke]

Solution

Thanks MFB. I think I got it. Here's the solution.S=1+4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}+4\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}I can find the sum of an Arithmetic-Geometric Series, hence\sum_{r=1}^{\infty}\frac{r}{5^{2r}}=\frac{25}{576}The Sum of third term, which is a simple Geometric Series,is\sum_{r=1}^{\infty}\frac{1}{5^{2r}}=\frac{1}{24}The Problem was the first term, which I happened to resolve thanks to MFB.

\sum_{r=1}^\infty \frac{r^2}{5^{2r}}=\frac{1}{25} \left(1+\sum_{r=1}^\infty \frac{2r+1}{5^{2r}}\right) + \frac{1}{25}\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}

\frac{24}{25}\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{25}\left(1+2\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}\right)\Rightarrow 4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{6}\left(1+2\times \frac{25}{576}+\frac{1}{24}\right)=\frac{325}{576}Substituting All the known variables into the First Equation,S=\frac{1728}{1728}+\frac{325}{1728}+\frac{300}{1728}+\frac{72}{1728}= \frac{2425}{1728}\Rightarrow \frac{1728}{485}S=\frac{1728}{485}\times \frac{2425}{1728}=5

If you do know calculus, that summation would be a standard homework...

...You can find $S(x)$ in closed form, so you can do all the calculations explicitly. Then, of course, you set x = 1/5.

While I do know Calculus, I can only solve them when there is that Integral Sign all over, or in physics. Using it in this manipulative way is not something I've done before. But still, if you could show me what you actually meant, I can learn something, as I need to brace myself for my College Entrance Exam, The IIT-JEE which is only a week apart. So, a little more help will be appreciated.

 

[Ray Vickson]

Thanks MFB. I think I got it. Here's the solution.S=1+4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}+4\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}I can find the sum of an Arithmetic-Geometric Series, hence\sum_{r=1}^{\infty}\frac{r}{5^{2r}}=\frac{25}{576}The Sum of third term, which is a simple Geometric Series,is\sum_{r=1}^{\infty}\frac{1}{5^{2r}}=\frac{1}{24}The Problem was the first term, which I happened to resolve thanks to MFB.


\frac{24}{25}\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{25}\left(1+2\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}\right)\Rightarrow 4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{6}\left(1+2\times \frac{25}{576}+\frac{1}{24}\right)=\frac{325}{576}Substituting All the known variables into the First Equation,S=\frac{1728}{1728}+\frac{325}{1728}+\frac{300}{1728}+\frac{72}{1728}= \frac{2425}{1728}\Rightarrow \frac{1728}{485}S=\frac{1728}{485}\times \frac{2425}{1728}=5


While I do know Calculus, I can only solve them when there is that Integral Sign all over, or in physics. Using it in this manipulative way is not something I've done before. But still, if you could show me what you actually meant, I can learn something, as I need to brace myself for my College Entrance Exam, The IIT-JEE which is only a week apart. So, a little more help will be appreciated.

What I mean is just exactly what I said:
\sum_{n=1}^{\infty} n x^n = x \frac{d}{dx} \sum_{n=1}^{\infty} x^n <br /> = x \frac{d}{dx} \frac{x}{1-x} = \frac{x}{(1-x)^2}
and
\sum_{n=1}^{\infty} n^2 x^n = x \frac{d}{dx} \frac{x}{(1-x)^2}.
You can just go ahead and do the derivative.

 

[AGNuke]

OK Thanks! Now I see... Thanks a bunch. I was forgetting the fact that S is the sum of an infinite converging GP.

 

[AGNuke]

Now then this is alternate solution to find that term which was bothering me, courtesy Ray Vickson
S=\sum_{r=1}^{\infty}x^{r}=\frac{x}{1-x}x\frac{\mathrm{d} S}{\mathrm{d} x}=\frac{x}{(1-x)^2}x\frac{\mathrm{d} }{\mathrm{d} x}\left (x\frac{\mathrm{d} S}{\mathrm{d} x} \right )=\frac{x(1+x)}{(1-x)^3}=\sum_{r=1}^{\infty}r^2x^r

Simply substituting $x=\frac{1}{25}$, I got $\frac{325}{1728\times 4}$, and multiplying it by 4 (as required) gives $\frac{325}{1728}$ which is the term I required.

Really Interesting using differentiation to find the summation so easily. I definitely will try this method in further question I will attempt.