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?

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 $\infty$.
