Show that : $\sum_{k=1}^{\infty}\frac{i^{k(5k+1)}}{k(k+1)}=1-\frac{π}{2}$

Question

Show that

$S=\displaystyle\sum_{k=1}^{\infty}\frac{i^{k(5k+1)}}{k(k+1)}=1-\frac{π}{2}$

My try :

$S=\displaystyle\sum_{k=1}^{\infty}\frac{e^{iπk(5k+1)/2}}{k(k+1)}$

$=\displaystyle\sum_{k=1}^{\infty}\frac{\cos (k(5k+1)π/2)}{k(k+1)}$

+$i\displaystyle\sum_{k=1}^{\infty}\frac{\sin (k(5k+1)π/2)}{k(k+1)}$

But I can't complete this because pass my level

Answer 1

Partial

We can easily notice that the pattern of signes is $(-,-,+,+,-,-,...)$ so our sum can be decomposed as follows:

$$\sum_{k=0}^{\infty} \frac{1}{(4k+3)(4k+4)}+\frac{1}{(4k+4)(4k+5)}-\sum_{k=0}^{\infty} \frac{1}{(4k+1)(4k+2)}+\frac{1}{(4k+2)(4k+3)}$$

So the problem can be reducted to:

$$\sum_{k=0}^{\infty} -\frac{8}{(4k+1)(4k+3)(4k+5)}$$

Answer 2

This equals the sum $$\sum_{k=1}^\infty\frac{(-1)^{k(5k+1)/2}}{k(k+1)}$$ Now the term $(-1)^{k(5k+1)/2}$ can be simplified as $$(-1)^{k(5k+1)/2}=\begin{cases}1&k\equiv0,3\mod{4}\\-1&k\equiv1,2\mod{4}\end{cases}$$ In other words the term follows the pattern $1,-1,-1,1,\dots$ and hence the sum can be written as $$-\frac1{1(1+1)}-\frac1{2(2+1)}+\frac1{3(3+1)}+\sum_{k=1}^\infty\left(\frac{1}{(4k)(4k+1)}-\frac{1}{(4k+1)(4k+2)}-\frac{1}{(4k+2)(4k+3)}+\frac{1}{(4k+3)(4k+4)}\right)$$ $$\begin{align} &=-\frac7{12}+\sum_{k=1}^\infty\left(\frac1{4k}-\frac1{4k+4}-\frac2{4k+1}+\frac2{4k+3}\right)\\ &=-\frac7{12}+\left(\frac14-\frac18\right)+\left(\frac18-\frac1{12}\right)+\left(\frac1{12}-\frac1{16}\right)+\dots\\ &+\left(-\frac25+\frac27\right)+\left(-\frac29+\frac2{11}\right)+\left(-\frac2{13}+\frac2{15}\right)+\dots\\ &=-\frac7{12}+\frac14-2\left(\frac15-\frac17+\frac19-\frac1{11}+\frac1{13}+\dots\right)\\ &=-\frac7{12}+\frac14-2\left(1-\frac13+\frac15-\frac17+\frac19-\frac1{11}+\frac1{13}+\dots\right)+2\left(1-\frac13\right)\\ &=-\frac7{12}+\frac14-2\left(\frac\pi4\right)+2-\frac23\\ &=1-\frac\pi2\\ \end{align}$$