# $\sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$ Computing this sum

Compute the limit:

$$\sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$$

My teacher says it can be solved by only using high school knowledge, but I don't see how. What did I try? Well, I thought of Riemann sums but I see no way to connect this sum to it. Thanks.

I'm only interested in a solution at high school level if possible !

UPDATE: Now, I'm my own teacher.

-
I cleared all the comments here, because discussion had devolved and almost nothing was relevant. Everyone, please maintain composure when commenting. –  mixedmath Sep 10 '12 at 3:59

A possible solution:

$\displaystyle \sum\limits_{k=1}^{+ \infty} \ln \left( 1+ \frac{1}{4k^2} \right)= \ln \left( \prod\limits_{k=1}^{+ \infty} \left( 1+ \frac{1}{4k^2} \right) \right)$. Using Euler's formula, $\displaystyle \prod\limits_{k=1}^{+ \infty} \left( 1+ \frac{1}{4k^2} \right)= -i\frac{2}{\pi} \sin \left( i \frac{\pi}{2} \right)= \frac{2}{\pi} \sinh \left( \frac{\pi}{2} \right)$.

Finally, $\displaystyle \sum\limits_{k=1}^{+ \infty} \ln \left(1+ \frac{1}{4k^2} \right)= \ln \left( \frac{2}{\pi} \sinh \left( \frac{\pi}{2} \right) \right)$.

-
Wallis formula is for the products of terms $1-1/(4k^2)$, and the problem has $1+1/(4k^2)$. Moreover, $\ln(2/\pi)<0$, while the sum is clearly positive. –  Julián Aguirre Sep 6 '12 at 10:21
Exactly. You cannot apply Wallis formula. –  Chris's sis Sep 6 '12 at 10:24
Exact. There is an analogous formula for complex numbers, so I edited my answer. Thank you. –  Seirios Sep 6 '12 at 10:52
solution is correct. –  Seyhmus Güngören Sep 6 '12 at 11:03

Take $f(x)=\ln\left(1+\frac{1}{4x^2}\right)$ over $[1,+\infty)$ and see that $f'(x)=-\frac{1}{2x^3(1+1/4x^2)}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $\displaystyle{\sum_{k=1}^{+\infty}\ln\left(1+\frac{1}{4k^2}\right)}$:

$$\int_1^{+\infty}\ln\left(1+\frac{1}{4x^2}\right)dx$$

Using integral by parts, you have the following answer as an upper bound:

$$\int_1^{+\infty}\ln\left(1+\frac{1}{4x^2}\right)dx=[-2\ln(2)x+x\ln(4+1/x^2)+\arctan(2x)]\big|_{x=1}^{+\infty}\\=2\ln(2)-\arctan(2)-\ln(5)+(1/2)\pi$$

For solving the integral take $\displaystyle{u=\ln\left(1+\frac{1}{4x^2}\right)}$ and $dv=dx$. :-)

-
So the series converges - that also follows from $$\log(1+x)\leq x\qquad (x>0)$$ –  AD. Sep 6 '12 at 11:09
@AD.: Yes I just could find an upper bound for his series not to find the exact value. I vote for Seirios. –  Babak S. Sep 6 '12 at 11:32
$\quad \ddot\smile \quad$ –  amWhy Mar 16 '13 at 0:49

The sum equals $$\ln\Bigl(\frac{2}{\pi}\sinh\Bigl(\frac{\pi}{2}\Bigr)\Bigr)=\ln\Bigl(\frac{e^{\pi/2}-e^{-\pi/2}}{\pi}\Bigr).$$ It can be obtained evaluating at $z=i\pi/2$ the product formula for the sine function $$\sin z=z\,\prod_{k=1}^\infty\Bigl(1-\frac{z^2}{\pi^2n^2}\Bigr),\qquad z\in\mathbb{C}.$$ I doubt that there is an elementary way to prove it.

-
agree! what is an elementary way to solve the problem, I also wonder. I also wonder which high school teaches it)) –  Seyhmus Güngören Sep 6 '12 at 11:04
@Julián, observa que utilizando el commando "\left(" or "\right)" dentro del signo de dólares automáticamente te ajusta el tamaño de los paréntesis. –  DonAntonio Sep 6 '12 at 13:17

The high school students I know would solve it like this.

-

As an alternative to @Seiros solution, albeit far less elegant is to use $$\log\left(1+\frac{1}{4k^2}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{1}{4^{n+1} k^{2n+2}}$$

Then $$\begin{eqnarray} \sum_{k=1}^\infty \log\left(1+\frac{1}{4k^2}\right) &=& \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{\zeta(2n+2)}{4^{n+1}} = \sum_{n=1}^\infty \frac{\pi^{2n}}{2n} \frac{B_{2n}}{(2n)!} \\ &=& \sum_{n=1}^\infty \frac{\pi^{n}}{n} \frac{B_{n}}{(n)!} + \frac{\pi}{2} \\&=& \frac{\pi}{2} + \int_0^1 \left(\sum_{n=1}^\infty \pi^{n} t^{n-1}\frac{B_{n}}{(n)!}\right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^1 \left(\frac{\pi}{\mathrm{e}^{\pi t}-1} - \frac{1}{t} \right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^\pi \left(\frac{1}{\mathrm{e}^{u}-1} - \frac{1}{u} \right) \mathrm{d} u \\ &=& \frac{\pi}{2} + \left. \log\left(\frac{1-\mathrm{e}^{-u}}{u}\right) \right|_{0}^{\pi} = \frac{\pi}{2} + \log\left(\frac{1-\exp(-\pi)}{\pi}\right) = \log\left(\frac{2}{\pi} \cdot \sinh\frac{\pi}{2}\right) \end{eqnarray} \tag{1}$$

-