<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] The approximate value of sum of terms in $HP$ is $S_n\approx \frac{1}{d}\ln(\frac{2a+2(n-1)d}{2a-d})$ here $a$=reciprocal of first term  $(a_1=4,a_2=3)$ respectively. $d$=difference between the reciprocal of two terms $(d_1=3,d_2=2)$ respectively
Note that the two HPs  are $3+3(\frac{1}{4}+\frac{1}{7}+....)-2(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)$ 
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