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

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

• @PeterForeman if $k=2$ we have $i^{2\times 11}=-1$, so this is not true – Eureka May 6 at 22:04
• Hint: $i^2=-1$, $i^3=-i$, $i^4=1$. – Théophile May 6 at 22:04
• Most formulas for $pi$ would have -1 to some power or C(k,x)...but this one does not! – NoChance May 6 at 22:14

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}

• Thank you very much Sir can you tell me what's the term $(-1)^{n(3n-1)/2}$ – Kînan Jœd May 6 at 22:28
• I don't know what you mean. I haven't written that in my working. That looks similar to what is in the numerator of the summation. – Peter Foreman May 6 at 22:38
• Not in your working , jut question I need term of $i^{n(3n-1)}$ – Kînan Jœd May 6 at 22:42

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:

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