Does the series $\frac{2}{4-1}+\frac{4}{16-1}+\dots+\frac{2k}{4k^2-1}$ have a sum up to $\infty$?

Question

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?

Answer 1

You have $$\frac{2k}{4k^2-1} \sim \frac{1}{2k}$$

so the series diverges.

Answer 2

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 $\infty$.

Answer 3

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

$$\sum_{k=1}^n \left(\frac{2k}{4k^2-1}-\frac1{2k}\right)=A_{2n}-\frac{n}{2n+1}\to \log 2-\frac12$$

Answer 4

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 $x\to 1$, $\tanh ^{-1}(x)\to \infty$.