# The calculation of a series

Calculate the series $$\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.$$

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What have you tried so far? –  Argon Jun 1 '12 at 21:11
Eat your peas. (Seriously though, do make an effort to include your work and to ask a question rather than state a demand.) –  rschwieb Jun 1 '12 at 21:12
Hello Marcos. Do you know about accepting answers to your questions? –  Antonio Vargas Jun 1 '12 at 23:46

Hint:

$$\frac 1 {(4n+1)(4n+3)}=\frac 1 2\left(\frac{1}{4n+1}-\frac 1{4n+3}\right)$$

Then note that

$$\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}} = \frac{1}{{2\left( {2n+1} \right) + 1}} - \frac{1}{{2\left( {2n} \right) + 1}}$$

from where

$$\sum\limits_{n = 1}^m {\left( {\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}}} \right)} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + - \cdots + \frac{1}{{2(2m + 1) + 1}} - \frac{1}{{2(2m) + 1}}$$

then recall the series for $\tan ^{-1}x$.

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@Henry Done. ´´´ –  Pedro Tamaroff Jun 1 '12 at 21:38

Here is another take on this. It may not be the most simple, but I think some may find it interesting.

Let $\chi(n)$ be the unique nontrivial Dirichlet character modulo $4$, so that by using $\frac{1}{(4n+1)(4n+3)}=\frac{1}{2}\left(\frac{1}{4n+1}-\frac{1}{4n+3}\right)$ your series equals $$\frac{1}{2}\sum_{n=1}^\infty \frac{\chi(n)}{n}=\frac{1}{2}L(1,\chi).$$ Noting that $\chi(n)=\left(\frac{d}{n}\right)$ with $d=-4$, and that the quadratic form $x^2+y^2$ is the only class with discriminant $D=-4$, we see that by the class number formula $$L(1,\chi)=\frac{2\pi}{\omega(d)\sqrt{|d|}}=\frac{\pi}{4},$$ where $\omega(d)$ is the number of symmetries of the corresponding complex lattice. (In our case $\omega(d)=4$, because it is the Gaussian integers).

Thus the original series evaluates to $\frac{\pi}{8}$.

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Agreed, very interesting! The $\equiv 1, 3 \bmod 4$ part of the statement suggested that something was up. –  Dylan Moreland Jun 2 '12 at 17:44

We let

$$f(z)=\dfrac{1}{(4z+1)(4z+3)}$$

There are two poles of $f(z)$. They are $4z_0+3=0 \implies z_0=-\frac{3}{4}$ and $4z_1+1=0 \implies z_1=-\frac{1}{4}$.

Residue calculus tells us that

$$\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(\text{sum of residues of }\pi\cot(\pi z)f(z))$$

We calculate the residues of the poles ($b_0$ and $b_1$ for $z_0$ and $z_1$ respectively):

$$b_0=\operatorname {Res}_{z=z_0}=\lim_{z \to z_0}\frac{(z-z_0)\pi\cot (\pi z)}{(4z+1)(4z+3)}$$

Using L'Hopital's rule we have

$$b_0=\lim_{z \to z_0} \frac{\pi\cot (\pi z)-(z-z_0)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-3\pi/4)}{4(4)}=-\frac{\pi}{8}$$

Similarly, we have

$$b_1=\lim_{z \to z_1} \frac{\pi\cot (\pi z)-(z-z_1)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-\pi/4)}{4(4)}=-\frac{\pi}{8}$$

So

$$\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(-\frac{\pi}{8}-\frac{\pi}{8})=\frac{\pi}{4}\implies \sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=\frac{1}{2}\frac{\pi}{4}=\frac{\pi}{8}$$

QED

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Once you splitted the initial fraction within the sum as below:

$$\frac 1 {(4n+1)(4n+3)}=\frac 1 2\left(\frac{1}{4n+1}-\frac 1{4n+3}\right)$$

then you may consider the following formula that is very helpful:

If $a+b+c+d=0$, $$\sum_{k=0}^\infty \left(\dfrac{a}{4k+1} + \dfrac{b}{4k+2}+\dfrac{c}{4k+3}+\dfrac{d}{4k+4}\right) = \dfrac{a-c}{8} \pi + \dfrac{a+c-2d}{4} \ln(2)$$

Replace the specific values you have and you're done. The limit is $\frac{\pi}{8}$.

The proof is complete.

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I'm interested in the formula you used. Can you give a few words about where it comes from? Or perhaps a reference? –  ranousse Jun 2 '12 at 18:28
@ranousse: you may want to see this math.stackexchange.com/questions/149120/… –  Chris's sis Jun 2 '12 at 19:02

$\frac{ \pi }{8}$ is what i get...

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Nice, perhaps you'd tell us about the "how". –  Gigili Jun 1 '12 at 21:20
Please don't downvote the guy, give him a chance. –  Pedro Tamaroff Jun 1 '12 at 21:25
Looking at the question and all of the effort shown by the OP in this thread, this appears to me to be a pretty appropriate answer. –  Jonas Meyer Jun 2 '12 at 6:00
@Jonas: I really like that interpretation. GIGO in this case becomes WIWO (work in work out). I fully support this standard. –  mixedmath Jun 7 '12 at 0:57

Here is another method.

We have that $$\sum_{n=0}^{\infty} x^{4n} = \dfrac1{1-x^4}$$ Integrate the above from $x=0$ to $x=t < 1$, to get $$\sum_{n=0}^{\infty} \dfrac{t^{4n+1}}{4n+1} = \int_0^{t} \dfrac{dx}{1-x^4} = \dfrac12 \left( \int_0^{t} \dfrac{dx}{1+x^2} + \int_0^{t} \dfrac{dx}{1-x^2} \right)\\ = \dfrac12 \left( \arctan(t) + \dfrac12 \left( \int_0^t \dfrac{dx}{1+x} + \int_0^t \dfrac{dx}{1-x} \right)\right)\\ =\dfrac12 \left( \arctan(t) + \dfrac12 \left( \log(1+t) - \log(1-t)\right)\right)$$ Now multiply throughout by $t$ to get, $$\sum_{n=0}^{\infty} \dfrac{t^{4n+2}}{4n+1} =\dfrac12 \left( t\arctan(t) + \dfrac12 \left( t\log(1+t) - t\log(1-t)\right)\right)$$ Now integrate the above from $t=0$ to $1$. Note that $$\int_0^1 t\arctan(t) dt = \dfrac{\pi-2}{4}$$ $$\int_0^1 t\log(1+t) dt = \dfrac14$$ $$\int_0^1 t\log(1+t) dt = -\dfrac34$$ The above integrals can be evaluated with relative ease by integration by parts. Hence, we now get that $$\sum_{n=0}^{\infty} \dfrac1{(4n+1)(4n+3)} = \dfrac12 \left( \dfrac{\pi-2}{4} + \dfrac12 \left( \dfrac14 - \left( - \dfrac34\right)\right)\right) = \dfrac{\pi}8$$

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