Calculate $ \sum_{n=0}^{∞}\frac{(-1)^n}{3n+2} $

Question

I'm having trouble solving this sum. Can't figure out what I'm dong wrong.

Thanks!

I'd start with having definite integrals, so plesae put limits there — SBF, Apr 20 '23 at 11:38
Please do not rely on pictures of text. Here's a MathJax tutorial :) — Shaun, Apr 20 '23 at 11:39
Also, while there is no problem with it here, it's always better to justify the exchange of limits. See https://math.stackexchange.com/questions/83721/when-can-a-sum-and-integral-be-interchanged With the correct integration bounds you will find the answer. Note that your primitive is not $0$ for $x=0$. — Jean-Claude Arbaut, Apr 20 '23 at 11:45
https://www.wolframalpha.com/input?i=int_0%5E1+x%2F%281%2Bx%5E3%29 I am EEE student — Bob Dobbs, Apr 20 '23 at 11:55
$\left[\arctan\frac{2x-1}{\sqrt3}\right]0^1=\frac\pi3\ne\frac\pi6=\left.\arctan\frac{2x-1}{\sqrt3}\right|{x=1}.$ — Anne Bauval, Apr 20 '23 at 12:12

score 1 · Answer 1 · answered Apr 20 '23 at 13:14

$$ \left[-\dfrac{\ln\left(\left|x+1\right|\right)}{3}+\dfrac{\ln\left(x^2-x+1\right)}{6}+\dfrac{\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}} \right]_{0}^{1}\\ = \dfrac{{\pi}}{2{\cdot}3^\frac{3}{2}}-\dfrac{6\ln\left(2\right)-\sqrt{3}{\pi}}{18} \approx 0.3735507278914242 $$

score 1 · Answer 2 · answered Apr 20 '23 at 13:54

The term $\displaystyle\left.\int f(x)\, dx \right|_{x=a}$ makes no sense since the antiderivative of a function is not unique, you need to consider the integration constant. If $\displaystyle\int f(x)\, dx = F(x)+C$ then $\displaystyle\left.\int f(x)\, dx \right|_{x=a} = F(a)+C$ is not determined until you specify $C$.

That's what definite integrals are for, you should use $\displaystyle\frac{1}{3n+2}=\int_{0}^1 x^{3n+1}\, dx$

Calculate $ \sum_{n=0}^{∞}\frac{(-1)^n}{3n+2} $

2 Answers2