How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$

Question

$$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$ 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.

Answer 1

$$\frac{1}{6(2n-1)(3n+1)}=\frac{1}{(6n-3)(6n+2)}=\frac{1}{5}\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)=\frac{1}{5}\int_0^1\left(x^{6n-4}-x^{6n+1}\right)$$

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.

Answer 2

Alternatively one may recall the series representation for the digamma function
$$\begin{equation} \psi(x+1) = -\gamma - \sum_{n=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n} \right), \quad \Re x >-1, \tag1 \end{equation} $$ where $\gamma$ is the Euler-Mascheroni constant.

Then by partial fraction decomposition we have $$ \begin{align} \frac1{(6n-3)(6n+2)} &= \frac15\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)\\\\ &=\frac1{30}\left[\left(\frac{1}{n-1/2}-\frac1n\right)-\left(\frac{1}{n+1/3}-\frac1n\right)\right] \end{align} $$ then summing from $n=1$ to $+\infty$, using $(1)$, we get $$ \sum_{n=1}^{\infty}\frac1{(6n-3)(6n+2)} =\frac1{30}\left(\psi\left(\frac43\right)-\psi\left(\frac12\right)\right) $$ equivalently

$$ \sum_{n=1}^{\infty}\frac1{(6n-3)(6n+2)} =\frac{1}{180} \left(18-\sqrt{3} \pi -9 \ln 3+24 \ln 2\right) $$

where we have used special values of the digamma function.

Answer 3

From this answer, we see that $$ H_{-1/2}=-2\log(2)\tag{1} $$ and $$ H_{-2/3}=-\frac32\log(3)-\frac\pi{2\sqrt3}\tag{2} $$ Therefore, $$ H_{1/3}=3-\frac32\log(3)-\frac\pi{2\sqrt3}\tag{3} $$ Applying $(1)$ and $(3)$ $$ \begin{align} \sum_{k=1}^\infty\frac1{6(2k-1)(3k+1)} &=\sum_{k=1}^\infty\frac1{30}\left(\frac2{2k-1}-\frac3{3k+1}\right)\\ &=\frac1{30}\sum_{k=1}^\infty\left(\frac1k-\frac1{k+\frac13}\right) -\frac1{30}\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\frac12}\right)\\ &=\frac1{30}\left(H_{1/3}-H_{-1/2}\right)\\[6pt] &=\frac1{30}\left(3-\frac32\log(3)-\frac\pi{2\sqrt3}+2\log(2)\right)\\[6pt] &=\frac{18-9\log(3)-\pi\sqrt3+12\log(2)}{180}\tag{4} \end{align} $$