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?

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?
