Does the series $\frac{2}{4-1}+\frac{4}{16-1}+\dots+\frac{2k}{4k^2-1}$ have a sum up to $\infty$? If there's one, then how do I find the sum?
I tried to rewrite the common term into partial fractions to see if it's a telescoping series but got a dead end there. How can I proceed now?
 A: You have
$$\frac{2k}{4k^2-1} \sim \frac{1}{2k}$$
so the series diverges.
A: If you know Taylor series, one thing which could be interesting is to notice that
$$\sum_{k=1}^\infty \frac {2k}{4k^2-1}x^{2k}=\frac{1}{2} x \tanh ^{-1}(x)+\frac{\tanh ^{-1}(x)}{2 x}-\frac{1}{2}$$ and when $x\to 1$, $\tanh ^{-1}(x)\to \infty$.
A: To expand on @TheSilverDoe's answer...
The sum is approximately $$\sum_{k=1}^\infty \frac{1}{2k}.$$
Note that this is $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\dots$$
Group the sum as follows:
$$\left(\frac{1}{2}\right)+\left(\frac{1}{4}\right)+\left(\frac{1}{6}+\frac{1}{8}\right)+\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}\right)+\dots$$
$$\frac{1}{2}\ge\frac{1}{4}$$
$$\frac{1}{4}\ge\frac{1}{4}$$
$$\frac{1}{6}+\frac{1}{8}\ge\frac{1}{4}$$
$$\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}\ge\frac{1}{4}$$
And so on. The reason for the last two is because $\frac{1}{6}>\frac{1}{8}$, and $2\cdot \frac{1}{8} = \frac{1}{4}$. Same logic for the next one, and on and on, for infinity.
Therefore, the series is divergent and leads to $\infty$.
A: As noticed the series diverges, more in detail we have that
$$\frac{2k}{4k^2-1}=\frac12\left(\frac{1}{2k-1}+\frac{1}{2k+1}\right)$$
therefore
$$\sum_{k=1}^n \frac{2k}{4k^2-1}=\frac12\sum_{k=1}^n \left(\frac{1}{2k-1}\right)+\frac12\sum_{k=1}^n \left(\frac{1}{2k+1}\right)=$$
$$=\frac12\sum_{k=1}^n \left(\frac{1}{2k-1}\right)+\frac12\sum_{k=2}^{n+1} \left(\frac{1}{2k-1}\right)=$$
$$=\frac12\sum_{k=1}^n \left(\frac{1}{2k-1}\right)+\frac12\sum_{k=1}^{n} \left(\frac{1}{2k-1}\right)+\frac12\frac1{2n+1}-\frac12=$$
$$=\sum_{k=1}^n \left(\frac{1}{2k-1}\right)-\frac{n}{2n+1}$$
and since
$$\sum_{k=1}^n \left(\frac{1}{2k-1}\right)=\frac12\sum_{k=1}^{n}\frac1k+\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}$$
we have
$$\sum_{k=1}^n \frac{2k}{4k^2-1}=\frac12\sum_{k=1}^{n}\frac1k+\sum_{k=1}^{2n} \frac{(-1)^{k+1}}{k}-\frac{n}{2n+1}=\frac12H_n+A_{2n}-\frac{n}{2n+1}\to \infty$$
and
$$\sum_{k=1}^n \left(\frac{2k}{4k^2-1}-\frac1{2k}\right)=A_{2n}-\frac{n}{2n+1}\to \log 2-\frac12$$
