Can anyone help me to find what is the value of $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ when it tends to infinity
The first i wanna find the pattern but it seems do not have any unique pattern can anyone help me?
On
the series equal to $$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}$$ the first term $$\frac{1}{1-x^6}=\sum_{n=0}^{\infty} x^{6n}$$ $$\int_0^1\frac{1}{1-x^6}dx=\int_0^1\sum_{n=0}^{\infty}x^{6n}dx$$ $$\int_0^1\frac{1}{1-x^6}dx=\sum_{n=0}^{\infty}\frac{1}{6n+1}$$
the second term
$$\frac{1}{1-x^6}=\sum_{n=0}^{\infty}x^{6n}$$ mutiply by $x^3$
$$\frac{x^3}{1-x^6}=\sum_{n=0}^{\infty}x^{6n+3}$$ $$\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\sum_{n=0}^{\infty}x^{6n+3}dx$$ $$\int_0^1\frac{x^3}{1-x^6}dx=\sum_{n=0}^{\infty}\frac{1}{6n+4}$$
hence $$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}=\int_0^1\frac{1}{1-x^6}dx-\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\frac{1}{1+x^3}dx$$ $$=\frac{1}{3}\log 2+\frac{\pi\sqrt{3}}{9}$$
On
Look at the final answer. The final answer tells you this series cannot be calculated simply. Here I propose a solution although it is not a good solution:
$$\Sigma_{n=1}^\infty \frac1{(n+a)(n+b)}=\frac{H_a-H_b}{a-b}$$
Where $H_x$ is harmonic number of $x$.
(It can be replaced with digamma function too. Although $\psi_0(x)\neq H_x$, the result is remained the same.)
Computing $H_x$ is not easy too. It is as hard as calculation of integrals. But if you have its table, you can calculate the series.
In this case, if you know $H_{\frac16}$ and $H_{\frac23}$ you can solve this series (Odd condition!).
$$\Sigma_{n=0}^\infty \frac{(-1)^n}{(3n+1)}=\Sigma_{n=0}^\infty (\frac1{(6n+1)}-\frac1{(6n+4)})=\Sigma_{n=0}^\infty \frac3{(6n+1)(6n+4)}=1-\frac14+\frac3{36}\Sigma_{n=1}^\infty \frac1{(n+\frac16)(6n+\frac46)}=1-\frac14+\frac1{12}\times\frac{H_{\frac46}-H_{\frac16}}{\frac46-\frac16}=\frac34+\frac16(H_{\frac23}-H_{\frac16})$$
I need to know these two harmonic numbers (the big disadvantage of this method):
$$H_{\frac23}=\frac32+\frac\pi{2 \sqrt3}-\frac{3 \log 3}2$$
$$H_{\frac16}=6-\frac{\sqrt3 \pi}2-\frac{3 \log 3}2-\log4$$
$$H_{\frac23}-H_{\frac16}=-\frac92+\frac{2\pi \sqrt3}3+2\log 2$$
$$\Sigma_{n=0}^\infty \frac{(-1)^n}{(3n+1)}=\frac34+\frac16 (-\frac92+\frac{2\pi \sqrt3}3+2\log 2)$$
$$\therefore ~~~\Sigma_{n=0}^\infty \frac{(-1)^n}{(3n+1)}=\frac{\pi \sqrt3}9+\frac{\log2}3$$
Hint. Observe that the general term of your series is $$\frac{(-1)^n}{3n+1}. $$ Then you may write $$ \begin{align} \sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}&=\sum_{n=0}^{\infty}(-1)^n\int_0^1 x^{3n} dx\\\\ &=\int_0^1 \sum_{n=0}^{\infty}(-1)^nx^{3n} dx\\\\ &=\int_0^1 \frac{1}{1+x^3} dx\\\\ &=\int_0^1 \left(\frac{1}{3 (1+x)}+\frac{2-x}{3 \left(1-x+x^2\right)}\right) dx\\\\ &=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9} \end{align} $$
Some details. $$ \begin{align} \int_0^1 \frac{1}{1+x^3} dx&=\frac{1}{3}\int_0^1\!\frac{dx}{(1+x)}+\frac{1}{3}\int_0^1\!\frac{2-x}{(x-1/2)^2+3/4} dx\\\\ &=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2-u}{u^2+3/4} du \quad (u=x-1/2, \,dx=du)\\\\ &=\frac{1}{3}\ln 2+\frac{1}{3}\int_{-1/2}^{1/2}\!\frac{3/2}{u^2+3/4} du\\\\ &=\frac{1}{3}\ln 2+\int_{0}^{1/2}\!\frac{1}{u^2+3/4} du\\\\ &=\frac{1}{3}\ln 2+\left.\frac{2}{\sqrt{3}}\arctan \left( \frac{2}{\sqrt{3}}u\right)\right|_{0}^{1/2} \\\\ &=\frac{1}{3}\ln 2+\frac{\pi\sqrt{3}}{9} \end{align} $$ where we have used $\displaystyle \arctan \!\left(\! \frac{1}{\sqrt{3}}\!\right)\!=\frac{\pi}{6}$.