sum of series $frac{1}{1cdot 3}+frac{1}{4cdot 5}+frac{1}{7cdot 7}+frac{1}{10cdot 9}+cdots$

The sum of series

$$\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots$$

My attempt $$\displaystyle \sum^{n}_{k=1}\frac{1}{(3k-2)(2k+1)}=\frac{1}{7}\sum^{n}_{k=1}\bigg[\frac{3}{3k-2}-\frac{2}{2k+1}\bigg]$$

$$=\frac{1}{7}\sum^{n}_{k=1}\int^{1}_{0}\bigg(3x^{3k-3}-2x^{2k}\bigg)dx$$

$$=\frac{1}{7}\int^{1}_{0}\bigg(\sum^{n}_{k=1}3x^{3k-3}-2x^{2k}\bigg)dx$$

$$=\frac{1}{7}\int^{1}_{0}\bigg[\frac{3(1-x^{3n})}{1-x^3}-\frac{2(1-x^{2n+2})}{1-x^2}\bigg]dx$$

How do i solve it Help me please

Answer

Such integrals can be expressed by the digamma function. One of the formulae for the digamma function is

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

You have then
$$\int_0^1 \frac{1-x^{3n}}{1-x^3} dx = \int_0^1 \frac{1-t^{n}}{1-t} \frac13t^{-\frac23}dt = \int_0^1 \frac13 \Big(\frac{1-t^{n-\frac23}}{1-t}-\frac{1-t^{-\frac23}}{1-t}\Big)dt = \frac13\big(\psi(n+\frac13)-\psi(\frac13)\big)$$
$$\int_0^1 \frac{x^2(1-x^{2n)}}{1-x^2} dx = \int_0^1 \frac{t(1-t^{n})}{1-t} \frac12t^{-\frac12}dt = \int_0^1 \frac12 \Big(2\frac{1-t^{n+\frac12}}{1-t}-2\frac{1-t^{\frac12}}{1-t}\Big)dt = \frac12\big(\psi(n+\frac32)-\psi(\frac32)\big)$$
so
$$\sum_{k=1}^n \frac{1}{(3k-2)(2k+1)} = \frac17\big(\psi(n+\frac13)-\psi(n+\frac32)-\psi(\frac13)+\psi(\frac32)\big)$$
The same formula can be derived faster, using the fact that $$\psi(n+z)-\psi(z) = \sum_{k=0}^{n-1}\frac{1}{z+k}$$
so
$$\sum_{k=1}^{n}\frac{1}{k-\frac23}=\sum_{k=0}^{n-1}\frac{1}{k+\frac13} = \psi(n+\frac13)-\psi(\frac13)$$

$$\sum_{k=1}^{n}\frac{1}{k+\frac12}=\sum_{k=0}^{n-1}\frac{1}{k+\frac32} = \psi(n+\frac32)-\psi(\frac32)$$

If you want to calculate the infinite sum, you have
$$\begin{align} \sum_{k=1}^\infty \frac{1}{(3k-2)(2k+1)} &= \frac17 \int_0^1 \Big(\frac{3}{1-x^3} - \frac{2x^2}{1-x^2}\Big) dx =\\ &= \frac17\int_0^1 \Big(\big(\frac{1}{1-x}+\frac{x+2}{x^2+x+1}\big) - \big(-2 + \frac{1}{1-x}+\frac{1}{x+1}\big)\Big) dx =\\ &= \frac17 \int_0^1 \Big(\frac{x+2}{x^2+x+1} +2 - \frac{1}{x+1}\Big) dx \end{align}$$
and the last integral can be computed with standard methods for the integrals of rational functions:

$$\begin{align} \int_0^1 \frac{x+2}{x^2+x+1} dx &= \int_0^1 \frac{(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34} dx =^{t=\frac{2}{\sqrt{3}}(x+\frac12)} \\ &= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{t+\sqrt{3}}{t^2+1} dt = \\ &= \Big(\frac{1}{2}\log(t^2+1) + \sqrt{3}\arctan t\Big)\Big|_{t=1/\sqrt{3}}^{t=\sqrt{3}} = \\ &= \frac{1}{2}\log 3 + \frac{\pi\sqrt{3}}{6}\end{align}$$

$$\int_0^1\frac{1}{x+1} dx = \log(x+1)\big|_{x=0}^{x=1}= \log 2$$
so
$$\sum_{k=1}^\infty \frac{1}{(3k-2)(2k+1)} =\frac17 (2+ \frac{\pi\sqrt{3}}{6}+\frac{1}{2}\log 3 -\log 2)$$