A Seemingly Simple Summation This one is from Knuth's Concrete Math and it is seriously giving me a hard time. The problem is to evaluate the following sum:
$$\sum_{k=1}^n (-1)^k \dfrac{k}{4k^2-1}$$
I tried splitting up the sum into parts, as in $$\sum_{i=1}^{\lfloor n/2 \rfloor} \dfrac{2i}{16i^2-1} - \sum_{j=1}^{\lceil n/2 \rceil} \dfrac{2j-1}{4(2j-1)^2-1}$$ but this only seemed to confuse the matter even more. I wish I had more work to show but I'm really stuck!
 A: Hint: Note that $\frac{k}{4k^2-1}=\frac{1}{4}\left(\frac{1}{2k-1}+\frac{1}{2k+1}\right)$. Forget about the $\frac{1}{4}$ for a while.
Now write down the first few terms in expanded form and observe the telescoping. The $(-1)^k$ plays a crucial role in making "neighbours" in the expanded form cancel.
A: $\sum_{k=1}^n(-1)^k\frac{k}{4k^2-1} =\sum_{k=1}^n(-1)^k\frac{k}{(2k-1)(2k+1)}$
$=\sum_{k=1}^n(-1)^k(\frac{1}{4(2k-1)} + \frac{1}{4(2k+1)})$
$=\sum_{k=1}^n(\frac{(-1)^k}{4(2k-1)} + \frac{(-1)^k}{4(2k+1)})$
$=\sum_{k=1}^n\frac{(-1)^k}{4(2k-1)} +\sum_{k=1}^n \frac{(-1)^k}{4(2k+1)}$
The last sum can be determined with general knowledge of series that lacks me for the moment.
Hope this helps you!!
A: Consider the partial sum
$$
\begin{align}
\sum_{k=1}^{2n}(-1)^k\frac{k}{4k^2-1}
&=\frac14\sum_{k=1}^{2n}(-1)^k\left(\frac1{2k-1}+\frac1{2k+1}\right)\\
&=\frac14\sum_{k=1}^n\left(-\frac1{4k-3}-\frac1{4k-1}+\frac1{4k-1}+\frac1{4k+1}\right)\\
&=\frac14\sum_{k=1}^n\left(-\frac1{4k-3}+\frac1{4k+1}\right)\\
&=-\frac14\sum_{k=1}^n\frac1{4k-3}+\frac14\sum_{k=2}^{n+1}\frac1{4k-3}\\
&=-\frac14+\frac14\frac1{4n+1}
\end{align}
$$
Then take the limit as $n\to\infty$.
