# Finding the sum of $\sum_{n=1}^\infty {2 \over (4n-1)(4n+3)}$

I know this is very similar to this question, so I thought the answer should be $\frac \pi 4$ until I saw another question that is more like mine that doesn't have a good answer (or if it is I am having trouble understanding it), but it is 2 years old so I don't think it will be getting another answer soon.

Anyway, I was able to prove that this sum converges by the direct comparison test to $\sum_{n=1}^\infty{\frac 1 {n^2}}$ (thanks to the help in my previous question), but now to find the sum.

I tried to use partial sums

$${2 \over (4n-1)(4n+3)} = {2 \over 19(4n-1)} - {1 \over 2(4n+3)}.$$

If this is right I think I am stuck since $19(4n-1)$ is always odd, but $2(4n+3)$ is always even so telescoping won't work unless I am mistaken.

• I tried to use partial sums Not a bad idea, but doublecheck your calculations there. – dxiv Jan 21 '18 at 23:23
• It should have been $\frac{2}{(4n-1)(4n+3)}=\frac{1}{2(4n-1)}-\frac{1}{2(4n+3)}$ – rtybase Jan 21 '18 at 23:31
• you made a mistake in computing partial fractions – G Cab Jan 21 '18 at 23:33
• Wow, I apologize to the community this was not a good question. I will make sure to check my work before I post from now on – dumbitdownjr Jan 21 '18 at 23:37
• You may check your work using GNU Octave online. I've just read the manual page for polynomial operations and find out that your problem can be solve by typing [r,p,k,e] = residue(2, conv([4 -1][4 +3])), which means that your are finding the partial fraction expression for the numerator $2$ divided by the product of $4n-1$ and $4n+3$. This gives a sum of fractions $\dfrac{r(i)}{(n - p(i))^{e(i)}}$ and a polynomial k, where $i$ is array index. – GNUSupporter 8964民主女神 地下教會 Jan 21 '18 at 23:48

## 2 Answers

$$\frac{1/2}{4n-1} - \frac{1/2}{4n+3} = \frac{2}{(4n-1)(4n+3)}$$ Hence your sum is telescoping $$\frac{1/2}{3} - \frac{1/2}{7} + \frac{1/2}{7} - \frac{1/2}{11} + \frac{1/2}{11} - \frac{1/2}{15} + \cdots$$ and it's clear that the sum is $\frac{1}{6}$.

There is a trick to find those fractions with little effort ( i will post here in a more general form )

If we have only two terms in the denominator $$\frac{1}{(ak+b)(ak+b+a) }=\frac{1}{a}\frac{(ak+b+a) -(ak+b)}{(ak+b)(ak+b+a) }$$ $$=\frac{1}{a}\frac{1}{(ak+b) } -\frac{1}{sa}\frac{1}{ (ak+b+a) }.$$

A still more general case: Calculate the sum $$\sum^{n}_{k=0}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }.$$

Every time we have this kind of summation we can use the telescoping sum. To find the terms of telescoping sum we can add in the numerator the following difference: $(ak+b+sa)$ the greatest term" minus thelowest term" $(ak+b)$, and to not alter the fraction divide by $\frac{1}{sa}$ (which is the resultant difference of those terms). $$\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }=\frac{1}{sa}\frac{(ak+b+sa) -(ak+b)}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }$$ $$=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) } -\frac{1}{sa}\frac{1}{(ak+b+a)\ldots (ak+b+sa) }.$$

Take $f(k)=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) }.$ Then the sum is $$\frac{-1}{sa}\sum^{n}_{k=0} \left(f(k+1) -f(k)\right)$$ by the telescopic sum it's $$=\frac{-1}{sa}\left(f(n+1) -f(0)\right)=\frac{-1}{sa}\left(\frac{1}{(an+b)\ldots (an+b+(s-1)a) } -\frac{1}{(b)(b+a)\ldots (+b+(s-1)a) }\right).$$

• Thank you for this interesting trick. Although I don't think I am good enough to use this yet, I am sure that others will find this helpful. – dumbitdownjr Jan 23 '18 at 3:37
• Thank you, i hope it could help somebody xD In the case of your question the trick to find the partial fractions is $$\frac{1}{(4n-1)(4n+3)}=\frac{(4n+3)-(4n-1)}{4(4n-1)(4n+3)}=\frac{1}{4}\frac{4n+3}{(4n-1)(4n+3)} - \frac{1}{4}\frac{(4n-1)}{(4n-1)(4n+3)}$$ $$=\frac{1}{4}\frac{1}{(4n-1)} - \frac{1}{4}\frac{1}{(4n+3)}.$$ So multiplying by $2$ you have the decomposition $$\frac{2}{(4n-1)(4n+3)}=\frac{1}{2}\frac{1}{(4n-1)} - \frac{1}{2}\frac{1}{(4n+3)}.$$ – Renji Rodrigo Jan 23 '18 at 3:51