<s>[INST] sum of series $\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 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
 [/INST] (CONSIDERING INFINITE SUM)
We have $$I_n=\frac 37\int_0^1 \frac {1-x^{3n}}{1-x^3} dx-\frac 27\int_0^1 \left(\frac {1-x^{2n+2}}{1-x^2} +\frac {x^2-1}{1-x^2}\right) dx$$
Let the first integral be denoted by $I_{1,n}$ and the second be denoted by $I_{2,n}$
Hence for the sum we need $$\lim_{n\to\infty} \frac 37 I_{1,n}-\frac 27 I_{2,n}$$
Now note that using the substitution $x^3=t$ in $I_{1,n}$ we get $$I_{1,n}=\frac 13 \int_0^1 \left(\frac {1-t^{n-\frac 23}}{1-t} -\frac {1-t^{-\frac 23}}{1-t}\right)dt$$
Now using the Integral relation if Digamma function I.e. $$\psi(z+1)+\gamma=\int_0^1 \frac {1-x^z}{1-x}dx$$
We get $$I_{1,n}=\frac 13\left(\psi\left(n+\frac 13\right)-\psi\left(\frac 13\right)\right)$$
Similarly using the substitution $x^2=u$ and the same integral relation in $I_{2,n}$ we get $$I_{2,n}= \frac 12\left(\psi\left(n+\frac 32\right)-\left[\psi\left(\frac 12\right)+\frac {1}{1/2}\right]\right)$$
But using that $$\psi(z+1)=\psi(z)+\frac 1z$$ we have $$I_{2,n}= \frac 12\left(\psi\left(n+\frac 32\right)-\psi\left(\frac 32\right)\right)$$
Now taking the limit we see that $$\lim_{n\to\infty} I_n=\frac 17\left[\lim_{n\to\infty} \left(\psi\left(n-\frac 23\right)-\psi\left(n+\frac 32\right)\right)\right]+\frac {\psi\left(\frac 32\right)-\psi\left(\frac 13\right)}{7}$$
Now for large $n$; $\psi(n)\sim \ln n$ . Using this the limit inside square brackets turns $0$ 
Thus the sum is equal to $$\frac {\psi\left(\frac 32\right)-\psi\left(\frac 13\right)}{7}$$
Which can be simplified using known values of Digamma function.
Edit: You can use from Wikipedia that $$\psi\left(\frac 13\right)=-\frac {\pi}{2\sqrt 3}-\frac {3\ln 3}{2}-\gamma$$
And $$\psi\left(\frac 32\right)=2-2\ln 2-\gamma$$
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