Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
$$f(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
After derivation:
$$f'(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}$$
The problem here is that:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}=x^2-x^{29}+x^{104}-...$$
Is it possible to find the closed form for the last series? 
2) Using partial sums:
$$S_n=\sum_{k=0}^{n}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
Now, using the formula:
$$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}\Rightarrow$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
$$S_n=\frac{1}{3}-\frac{1}{30}+...+(-1)^{n}\frac{1}{(n+1)(4(n+1)^2-1)}$$ 
$$\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}=T_n=-\frac{1}{30}+...+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
$$T_n=S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Going back to the formula $$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
we have that $S_n$ cancels, so we can't determine partial sums using this method?
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+T_n$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Question: How to find the sum of this series?
 A: 
The way to proceed is to exploit the telescoping series as illustrated by @MarkusScheuer.  But, ... 

If one really wants to use the method proposed in the OP, then writing 
$$\frac{1}{k(4k^2-1)}=\int_0^1 x^{k(4k^2-1)-1}\,dx$$
won't lead to tractable way forward since $\sum_{k=1}^K x^{k(4k^2-1)-1}$ is not a geometric series.  
However, we can proceed by writing
$$\begin{align}
\frac{1}{k(4k^2-1)}&=\frac{1}{2k+1}+\frac{1}{2k-1}-\frac1k\\\\
&=\int_0^1 (x^{2k}+x^{2k-2}-x^{k-1})\,dx \tag 1
\end{align}$$
Then, using $(1)$ it is easy to see that
$$\begin{align}
\sum_{k=1}^K (-1)^{k-1}\frac{1}{k(4k^2-1)}&=-\int_0^1 \sum_{k=1}^K((-x^2)^k+x^{-2}(-x^2)^k-x^{-1}(-x)^k)\,dx\\\\
&=\int_0^1 (1 -(-1)^Kx^{2K}) \,dx-\int_0^1 \frac{1-(-1)^Kx^{K}}{1+x}\,dx \\\\
&=1-\frac{(-1)^K}{2K+1}-\log(2)+(-1)^K\int_0^1 \frac{x^K}{1+x}\,dx\tag 2
\end{align}$$
whence letting $K\to \infty$ in $(2)$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty (-1)^{k-1}\frac{1}{k(4k^2-1)}=1-\log(2)}$$

as expected!

NOTE:
There are a number of ways to show that the integral on the right-hand side of $(2)$ approaches $0$ as $K\to \infty$.  
One way is to apply the Dominated Convergence Theorem with dominating function $\frac{1}{1+x}$.
A second way is to integrate by parts with $u=\frac1{1+x}$ and $v=\frac{x^{K+1}}{K+1}$. 
A third way is to note that $0\le \int_0^1 \frac{x^K}{1+x}\,dx\le \int_0^1x^K\,dx=\frac{1}{K+1}$ and apply the squeeze theorem.
A: 
With telescoping we obtain
\begin{align*}
\sum_{k=1}^\infty& (-1)^{k-1}\frac{1}{k(4k^2-1)}\\
&=\lim_{N\to\infty}\sum_{k=1}^N(-1)^{k-1}\frac{1}{k(4k^2-1)}\\
&=\lim_{N\to\infty}\sum_{k=1}^N(-1)^{k-1}\left(\frac{1}{2k-1}+\frac{1}{2k+1}-\frac{1}{k}\right)\\
&=1-\sum_{k=1}^\infty(-1)^{k-1}\frac{1}{k}\\
&=1-\ln 2
\end{align*}

A: Observe that checking for absolute convergence we get
$$\frac1{k(4k^2-1)}\le\frac1{k^2}\;\implies\;\text{since}\;\;\sum_{k=1}^\infty\frac1{k^2}$$
converges so does the series of the left side. What can you then deduce?
A: An alternative approach: since $\frac{1}{k(4k^2-1)} = \frac{1}{2k-1}-\frac{2}{2k}+\frac{1}{2k+1}$ is the integral over $(0,1)$ of $x^{2k-2}-2x^{2k-1}+x^{2k} = x^{2k-2}(1-x)^2$, we have:
$$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{k(4k^2-1)}=\int_{0}^{1}(1-x)^2\sum_{k\geq 1}(-1)^{k+1} x^{2k-2}\,dx = \int_{0}^{1}\frac{(1-x)^2}{1+x^2}\,dx $$
and the last integral can be re-written as
$$ 1-\int_{0}^{1}\frac{2x}{1+x^2}\,dx = 1-\log(1+x^2)_{x=1} = \color{red}{1-\log 2}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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This is a variant of $\texttt{@Markus Scheuer}$ 'telescopic' fine answer:

\begin{align}
\sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k\pars{4k^{2} - 1}} & =
-2\sum_{k = 1}^{\infty}
{\ic^{2k} \over \pars{2k}\bracks{\pars{2k}^{2} - 1}} =
-2\,\Re\sum_{k = 2}^{\infty}{\ic^{k} \over k\pars{k^{2} - 1}}
\\[5mm] & =
-\,\Re\pars{\sum_{k = 2}^{\infty}{\ic^{k} \over k - 1} -
2\sum_{k = 2}^{\infty}{\ic^{k} \over k} +
\sum_{k = 2}^{\infty}{\ic^{k} \over k + 1}}
\\[5mm] & =\require{cancel}
-\,\Re\bracks{\cancel{\ic\sum_{k = 1}^{\infty}{\ic^{k} \over k}} +
2\ic - \color{#f00}{2\sum_{k = 1}^{\infty}{\ic^{k} \over k}} -
\color{#f00}{\ic}\pars{\color{#f00}{-\ic} + {1 \over 2} +
\cancel{\sum_{k = 1}^{\infty}{\ic^{k} \over k}}}}
\\[5mm] & =
2\,\Re\sum_{k = 1}^{\infty}{\ic^{k} \over k} + 1 =
-2\,\Re\ln\pars{1 - \ic} + 1 = \bbx{\ds{1 - \ln\pars{2}}}
\end{align}
