Calculate the series \begin{equation} \sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}. \end{equation}
The calculation of a series
-
0Hello Marcos. Do you know about [accepting answers](http://meta.stackexchange.com/a/5235) to your questions? – 2012-06-01
6 Answers
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 $\tan ^{-1}x$.
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 $\frac{\pi}{8}$.
-
0Agreed, very interesting! The $\equiv 1, 3 \bmod 4$ part of the statement suggested that _something_ was up. – 2012-06-02
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
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.
-
1@ranousse: you may want to see this http://math.stackexchange.com/questions/149120/is-this-sum-related-to-the-gregorys-limit – 2012-06-02
$\frac{ \pi }{8}$ is what i get...
-
1@Jonas: I really like that interpretation. [GIGO](http://en.wikipedia.org/wiki/Garbage_In,_Garbage_Out) in this case becomes WIWO (work in work out). I fully support this standard. – 2012-06-07
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 $\sum_{n=0}^{\infty} \dfrac1{(4n+1)(4n+3)} = \dfrac12 \left( \dfrac{\pi-2}{4} + \dfrac12 \left( \dfrac14 - \left( - \dfrac34\right)\right)\right) = \dfrac{\pi}8$