<s>[INST] Finding the value of the infinite sum $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ 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?
 [/INST] 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 
$$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}=\int_0^1\frac{1}{1-x^6}dx-\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\frac{1}{1+x^3}dx$$
$$=\frac{1}{3}\log 2+\frac{\pi\sqrt{3}}{9}$$
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