What does this series converge to? $\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$ What does the following expression converge to?
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$$
(It looks like this problem)
$\displaystyle\sum_{k=0}^n a_{2k+1}+a_{2k+2}-a_{k+1}$ ;
$\displaystyle\sum_{k=0}^n a_{k+1}$ does not converge.

$$\mathop {\lim }\limits_{n \to  + \infty } \sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$
$$=\mathop {\lim }\limits_{n \to  + \infty } \sum_{k=0}^n\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\mathop {\lim }\limits_{n \to  + \infty } \sum_{k=0}^n\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)=$$
$$=\frac{\pi}{8}+\frac{\ln2}{4}-\left(\frac{\pi}{8}-\frac{\ln2}{4}\right)=\frac{\ln2}{2}.$$
