Calculate the sum of the series $\sum_{n=1}^{+\infty} \frac{c_n}{n}$ 
Calculate the sum of the series $\sum_{n=1}^{+\infty} \frac{c_n}{n}$ for $$c_n=\begin{cases} 3, n=4k+1, k\in \mathbb N \\ -1, n\neq4k+1 \end{cases}$$

My solution:Assuming that $\mathbb N={1,2,...}$ we have $$\sum_{n=1}^{+\infty} \frac{c_n}{n}=-1+(-1\cdot 3)+3+(-1\cdot 3)+3+(-1\cdot 3)+3+...$$However "last word of $c_n$" is infinite so I don't know how I can finish this task because we see that $-3$ shortens periodically from $3$ but I think that I can't write that $\sum_{n=1}^{+\infty} \frac{c_n}{n}=-1$.Can you give your opinion on this subject?
 A: Note that the sum consists of blocks of $$\frac 3{4k+1}-\frac 1{4k+2}-\frac 1{4k+3}-\frac 1{4k+4}$$
If you put all these over a common denominator and simplify the $k^3$ terms in the numerator cancel and you are left with 
$$\frac{96k^2+136k+46}{(4k+1)(4k+2)(4k+3)(4k+4)}$$
Because this is of order $k^{-2}$ the sum will converge.  Alpha gets the sum to be $$\frac \pi 2 + \log(2)$$
Seeing the answer, I am tempted to write
$$\log(2)=\sum_{k=0}^\infty\frac 1{4k+1}-\frac 1{4k+2}+\frac 1{4k+3}-\frac 1{4k+4}$$
and try to get
$$\frac \pi 2=2\sum_{k=0}^\infty\frac 1{4k+1}-\frac 1{4k+3}=2\sum_{k=1}^\infty \frac 2{(4k+1)(4k+3)}$$
A: $$\begin{eqnarray*}&&\sum_{k\geq 0}\left[\frac{3}{4k+1}-\frac{1}{4k+2}-\frac{1}{4k+3}-\frac{1}{4k+4}\right]\\&=&\sum_{k\geq 0}\int_{0}^{1}(3-x-x^2-x^3)x^{4k}\,dx \\&=&\int_{0}^{1}\frac{1}{1-x^4}(3-x-x^2-x^3)\,dx\\&=&\int_{0}^{1}\frac{3+2x+x^2}{(1+x)(1+x^2)}\,dx\\&=&\int_{0}^{1}\frac{2(1+x)+(1+x^2)}{(1+x)(1+x^2)}\,dx =\frac{\pi}{2}+\log(2)\end{eqnarray*}$$
by partial fraction decomposition.
A: As Ross Millikan wrote, consider the partial sum
$$S_n=\sum_{k=1}^n\left(\frac 3{4k+1}-\frac 1{4k+2}-\frac 1{4k+3}-\frac 1{4k+4}\right)$$ and remember that
$$T(a,b)=\sum_{k=1}^n \frac 1{ak+b}=\frac{H_{n+\frac{b}{a}}-H_{\frac{b}{a}}}{a}$$ where appear harmonic numbers and that the expansion for large values of $n$ is given by
$$T(a,b)=\frac{\log \left({n}\right)+\gamma-H_{\frac{b}{a}} }{a}+\frac{a+2 b}{2 a^2
   n}+O\left(\frac{1}{n^2}\right)$$
So,
$$S_n=3T(4,1)-T(4,2)-T(4,3)-T(4,4)=\left(\frac{\pi }{2}+\log (2)-\frac{23}{12}\right)-\frac{3}{8n}+O\left(\frac{1}{n^2}\right)$$ which gives the limit and shows how it is approached.
