<s>[INST] 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.
 [/INST] We can write the sum like this:
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\sum_{k=0}^n\left(\frac{1}{4k+1}+\frac{1}{4k+2}+\frac{1}{4k+3}+\frac{1}{4k+4}\right)-\sum_{k=0}^n\left(\frac{1}{4k+4}+\frac{3}{2}\left(\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{2k+2}\right)\right)=H_{4n+4}-\frac{1}{4}H_{n+1}-\frac{3}{2}H_{2n+2}+\frac{3}{4}H_{n+1}=H_{4n+4}+\frac{1}{2}H_{n+1}-\frac{3}{2}H_{2n+2}$$
Using $H_n = \ln n+\gamma +\mathcal{E}_n$ with $\mathcal{E}_n \to 0$, we can see that:
$$\lim_{n\to \infty}\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\lim_{n\to \infty} \left(\ln(4n+4)+\frac{1}{2}\ln(n+1)-\frac{3}{2}\ln(2n+2)\right)=\ln \sqrt{2}$$
</s>