how to proof the convergence of this series?

Question

I need to prove the convergence of this series : $$\sum_{n=0}^\infty (-1)^n\biggl(\frac{1}{3n^2+4n+1}\biggl) = \frac{\pi}{\sqrt{12}}$$

Answer 1

The series is absolutely convergent: comparison with $\sum 1/n^2$. I presume you really wanted to ask why it converges to $\pi/\sqrt{12}$. Well, $$\frac2{3n^2+4n+1}=\frac{3}{3n+1}-\frac1{n+1}=3\left(\frac1{3n+1} -\frac1{3n+3}\right).$$ Your sum equals \begin{align} \frac32\sum_{n=0}^\infty(-1)^n \left(\frac1{3n+1}-\frac1{3n+3}\right) &=\frac32\sum_{n=0}^\infty(-1)^n\int_{0}^1(t^{3n}-t^{3n+2})\,dt\\ &=\frac32\int_0^1\frac{1-t^2}{1+t^3}\,dt\\ &=\frac32\int_0^1\frac{1-t}{1-t+t^2}\,dt. \end{align} This is an integral one can do via standard calculus methods.

Answer 2

the alternating series test consists of proving that $$|a_n|$$ is monotonous decreasing and $$\lim_{n\to \infty}a_n=0$$ the secont part is clear, for the first part we have $$a_{n+1}-a_n=-{\frac {6\,n+7}{ \left( n+1 \right) \left( 3\,n+1 \right) \left( n+ 2 \right) \left( 3\,n+4 \right) }} <0$$ for $n>0$