What does this series converge to? $\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$

Question

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.

Answer 1

Another way: $$\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$ $$=\sum_{k=0}^{+\infty}\left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right)-\sum_{k=0}^{+\infty}\left(\frac{1}{4k+2}-\frac{1}{4k+4}\right)=$$ $$=\ln2-\frac{\ln2}{2}=\frac{\ln2}{2}.$$

Answer 2

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

Answer 3

@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 $0.3459$.

Answer 4

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:

$$\lim_{n\to \infty}\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=\lim_{n\to \infty} \left(\ln(4n+4)+\frac{1}{2}\ln(n+1)-\frac{3}{2}\ln(2n+2)\right)=\ln \sqrt{2}$$