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.
 A: $$\mathop {\lim }\limits_{n \to  + \infty } \sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$
$$=\mathop {\lim }\limits_{n \to  + \infty } \sum_{k=0}^n\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\mathop {\lim }\limits_{n \to  + \infty } \sum_{k=0}^n\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)=$$
$$=\frac{\pi}{8}+\frac{\ln2}{4}-\left(\frac{\pi}{8}-\frac{\ln2}{4}\right)=\frac{\ln2}{2}.$$
A: @Michael Rozenberg gave the answer by telescoping.
We can also have more than the limit itself since
$$S_n=\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\frac{1}{2} \left(H_{2 n+\frac{3}{2}}-H_{n+\frac{1}{2}}\right)$$ where appear harmonic numbers.
Using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^3}\right)$$ apply it twice and continue with Taylor series to get
$$S_n=\frac{\log (2)}{2}-\frac{1}{64 n^2}+O\left(\frac{1}{n^3}\right)$$
Computing for $n=5$
$$S_5=\frac{1170028}{3380195}\approx 0.3461$$ while the truncated series gives $0.3459$.
A: 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:
$$\lim_{n\to \infty}\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\lim_{n\to \infty} \left(\ln(4n+4)+\frac{1}{2}\ln(n+1)-\frac{3}{2}\ln(2n+2)\right)=\ln \sqrt{2}$$
A: Another way:
$$\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$
$$=\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right)-\sum_{k=0}^{+\infty}\left(\frac{1}{4k+2}-\frac{1}{4k+4}\right)=$$
$$=\ln2-\frac{\ln2}{2}=\frac{\ln2}{2}.$$
