Sum of Gaps 2

Since $\sum_{k=1}^n (-1)^{k+1}x^{2(k-1)} \; = \; \frac{1 - (-x^2)^n}{1+x^2} \hspace{2cm} 0 < x < 1$ we deduce that $\sum_{k=1}^n \frac{(-1)^{k+1}}{2k-1} \; = \; \int_0^1 \frac{1 - (-x^2)^n}{1+x^2} \,dx \; = \; \tfrac14\pi - (-1)^n\int_0^1 \frac{x^{2n}}{1+x^2}\,dx$ so that $\pi - 4\sum_{k=1}^n \frac{(-1)^{k+1}}{2k-1} \; = \; 4(-1)^n \int_0^1 \frac{x^{2n}}{1+x^2}\,dx$ for all integers $n \ge 1$ . Thus $S_N \; = \; \sum_{n=1}^N\left(\pi - 4\sum_{k=1}^n \frac{(-1)^{k+1}}{2k-1} \right) \; = \; 4\sum_{n=1}^N (-1)^n \int_0^1 \frac{x^{2n}}{1+x^2}\,dx \; = \; -4\int_0^1\frac{x^2(1 - (-x^2)^{N})}{(1+x^2)^2}\,dx$ from which it is clear that the infinite sum is $S \; = \; -4\int_0^1 \frac{x^2}{(1 + x^2)^2}\,dx$ The substitution $x = \tan\theta$ gives $S \; = \; -4\int_0^{\frac14\pi}\sin^2\theta\,d\theta \; = \; -2\int_0^{\frac14\pi}(1 - \cos2\theta)\,d\theta \; = \; \boxed{1 - \tfrac12\pi}$

https://photos.app.goo.gl/s44ckcDS8bW5b8Qf6

If someone (mod) can convert it to a solution that would be a lot helpful.

I’d highly suggest checking out Desmos or Wolfram Alpha.

Adding up consecutive terms, we see that the $(2n)$ th gap is $\sum_{k=n}^{\infty}\left(\frac{4}{4k+1}-\frac{4}{4k+3}\right)=\sum_{k=n}^{\infty}\frac{8}{(4k+1)(4k+3)}$ . Likewise, the $(2n-1)$ th gap is $-\sum_{k=n}^{\infty}\frac{8}{(4k+1)(4k-1)}$ . The sum of these two gaps is $-\sum_{k=n}^{\infty}\frac{32}{(4k-1)(4k+1)(4k+3)}$ , and the grand total of all gaps is $-\sum_{k=n}^{\infty}\frac{32k}{(4k-1)(4k+1)(4k+3)}=\frac{1}{2}(2-\pi)\approx \boxed{-0.571}$ .

\sum _{n=1}^{\infty } \left(\pi -4 \sum _{k=1}^n \frac{(-1)^{k+1}}{2 k-1}\right)

is \sum _{n=1}^{\infty } 2 (-1)^n \Phi \left(-1,1,n+\frac{1}{2}\right)

Using Wolfram Mathematica 11.3.

Evaluating that expression numerically to machine precision, the answer is -0.570796326795 which is the answer given by Mark Hennings

Used Double Sum in WolframAlpha

(1) -0.5658 with 100 terms

(2) -0.5683 with 200 terms

(3) -0.5688 with 250 terms

(4) -0.5698 with 500 terms

(5) -0.57017 with 800 terms

Answer = -0.570

Far to be as elegant as Mark Hennings solution,

Just compute the formulas using a programming language

1
2
3
4
5
6

import math
c,s=0,0
for a in range(1, 10**6): #10^6 instead of infinity
    c+=(-1 if a%2==0 else 1)/(2*a-1)
    s+=(math.pi-4*c)
print(s)

output: -0.5707967703631267

Hence the answer is aproximatively $\boxed{-0.5708}$