INFINITE SERIES RELATED TO SPECIAL VALUES OF HARMONIC NUMBERS

INFINITE SERIES RELATED TO SPECIAL VALUES OF HARMONIC NUMBERS

From this Twitter post I saw the following results.

$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$ (1)

$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$ (2)

$\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 \qquad \tag{3}$ (3)

$\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 \qquad \tag{4}$ (4)

These sums above are related to the Digamma function and the Harmonic numbers as we will see below

Recall the harmonic number

$H_n=\sum_{k=1}^{n}\frac{1}{k}$

$=\sum_{k=1}^{n}\int_0^1 x^{k-1}dx$

$=\int_0^1 \sum_{k=1}^{n} x^{k-1}dx$

$H_n=\int_0^1\frac{1-x^n}{1-x}dx$ (5)

From (5) we can generalize the Harmonic numbers to non integer numbers switching $n \mapsto x$

$H_x=\int_0^1\frac{1-t^x}{1-t}dt$ (6)

Now recall the digamma function and it´s infinite series representation

$\psi(x+1)=-\gamma+\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$ (7)

it´s integral representation

$\psi(x+1)=-\gamma+\int_0^1\frac{1-t^x}{1-t}dt$ (8)

and it´s recurrence equation

$\psi(x+1)=\frac{1}{x}+\psi(x)$ (9)

From (6) and (8) we get that

$\psi(x+1)=-\gamma+H_x$ (10)

and from (9) we obtain

$H_x=\gamma+\frac{1}{x}+\psi(x)$ (11)

From (7) we get

$\psi(x)+\frac{1}{x}+\gamma=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+x}$ (12)

With equation (12) above and the special values of the Digamma function proved here, We are now equipped to compute the sums (1) to (4)

For (1) by partial fractions

$\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{2}{2n+1}$

$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}$

From (11) and (12) we obtain

$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=\psi\left(\frac{1}{2}\right)+2+\gamma$ (13)

$\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{2}}=-\gamma -2\ln 2+2+\gamma$

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2}$ (14)

Similarly (2)

$\sum_{n=1}^{\infty}\frac{1}{n(3n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{3}{3n+1}$

$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{3}}=\psi\left(\frac{1}{3}\right)+3+\gamma$

$=-\frac{\pi}{2 \sqrt{3}}-\frac{3 \ln 3}{2}-\gamma+3+\gamma$

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}}}$ (15)

Next

$\sum_{n=1}^{\infty}\frac{1}{n(4n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{4}{4n+1}$

$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{4}}=\psi\left(\frac{1}{4}\right)+4+\gamma$

$=-\frac{\pi}{2}-3 \ln 2-\gamma+4+\gamma$

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n(4 n+1)}=4-\frac{\pi}{2}-3 \ln 2 }$ (16)

Finally

$\sum_{n=1}^{\infty}\frac{1}{n(6n+1)}=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{6}{6n+1}$

$=\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{n+\frac{1}{6}}=\psi\left(\frac{1}{6}\right)+6+\gamma$

$=-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2-\gamma +6+\gamma$

$\boxed{\sum_{n=1}^{\infty} \frac{1}{n(6 n+1)}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 }$ (17)

From (11) and (13) we obtain

$\boxed{H_{\frac{1}{2}}=2-2\ln 2}$ (18)

From (11) and (15) we obtain

$\boxed{H_{\frac{1}{3}}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}}}$ (19)

From (11) and (16) we obtain

$\boxed{H_{\frac{1}{4}}=4-\frac{\pi}{2}-3 \ln 2 }$ (20)

From (11) and (17) we obtain

$\boxed{H_{\frac{1}{6}}=6-\frac{\sqrt{3} \pi}{2}-\frac{3 \ln 3}{2}-2 \ln 2 }$ (21)

The special values above can be seen here

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