<s>[INST] How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$ $$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the squeeze theorem by putting the $a_n$ between $\frac{1}{(2n-1)(2n+1)}$ and $\frac{1}{(4n-1)(4n+1)}$ but it doesn't work.
 [/INST] $$\frac{1}{6(2n-1)(3n+1)}=\frac{1}{(6n-3)(6n+2)}=\frac{1}{5}\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)=\frac{1}{5}\int_0^1\left(x^{6n-4}-x^{6n+1}\right)$$
by using partial fractions and noting that $\int_0^1x^a\,dx=\frac{1}{a+1}$. Then:
$$\sum_{n=1}^N\frac{1}{6(2n-1)(3n+1)}=\frac{1}{5}\int_0^1\sum_{n=1}^N\left(x^{6n-4}-x^{6n+1}\right)=\frac{1}{5}\int_0^1(x^2-x^7)\frac{1-x^{6N}}{1-x^6}\,dx$$
and this last integral is equal to:
$$\frac{1}{5}\int_0^1(x^2-x^7)\frac{1}{1-x^6}\,dx-\frac{1}{5}\int_0^1\frac{x^{6N+2}(1-x^5)}{1-x^6}\,dx$$
Upon cancelling fractions, the first of these is equal to $\frac{1}{5}\int_0^1x^2\frac{1+x+x^2+x^3+x^4}{1+x+x^2+x^3+x^4+x^5}\,dx$, and the latter can be shown to be convergent to $0$ by noting that $\frac{1-x^5}{1-x^6}\le1$ on the given interval and $\int_0^1x^{6N+2}\,dx\to0$.
So, the series converges to $\frac{1}{5}\int_0^1x^2\frac{1+x+x^2+x^3+x^4}{1+x+x^2+x^3+x^4+x^5}\,dx$.
This integral can be evaluated exactly, if needed, by expanding the integrand into partial fractions.
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