$\sum_{n=1}^\infty \log \frac{4n^2+1}{4n^2}$ I'm stuck with the following series:
$$\sum_{n=1}^\infty \log \frac{4n^2+1}{4n^2}$$
It appears to converge, however I've not been able to prove it. Any clues?
Also, provided the convergence, any method to compute the sum?
 A: $$S = \sum_{n=1}^\infty \log \frac{4n^2+1}{4n^2}$$
$$e^{S} = \prod_{n=1}^\infty \left (1+\frac{1}{4n^2} \right )$$
$$e^{-S} = \prod_{n=1}^\infty \left (\frac{4n^2}{1+4n^2} \right )$$
Notice that $$\frac{\pi}{2} = \prod_{n=1}^\infty \left (\frac{4n^2}{4n^2-1} \right ),$$
this is the famous Wallis product.
So the product stemming from your equation must converge since the denominator of the product for $e^{-S}$ is bigger. Hence your sum is convergent.
The Wallis product is a specific case of $$ \frac{\sin(\pi z)}{\pi z} = \prod_{n=1}^\infty \left (1-\frac{z^{2}}{n^{2}} \right )$$
so just pick $z = \frac{i}{2}$ to get the answer.
A: Just for convergence,
this elementary inequality
is enough:
For $z > 0$,
$\dfrac1{z+1}
\lt \ln(1+\frac1{z})
\lt \dfrac1{z}
$.
Then,
since
$\log \frac{4n^2+1}{4n^2}
=\log(1+\frac1{4n^2})
\lt \frac1{4n^2}
$,
the sum of these converges.
Here is the elementary proof
of the inequality.
For $x > 0$,
$\begin{array}\\
\ln(1+x)
&=\int_0^x \dfrac{dt}{1+t}\\
&\lt\int_0^x dt\\
&= x\\
\text{and}\\
\ln(1+x)
&=\int_0^x \dfrac{dt}{1+t}\\
&\gt\int_0^x \dfrac{dt}{1+x}\\
&= \dfrac{x}{1+x}\\
\text{so}\\
\ln(1+\frac1{z})
&> \dfrac{1/z}{1+1/z}\\
&= \dfrac{1}{z+1}\\
\end{array}
$
For this problem
you do not need the
second inequality,
but it useful when
you want to prove divergence.
