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How would you prove convergence/divergence of the following series?

$$\sum\limits_{n\geq 1}\dfrac{1}{(3n-1)(3n+1)}$$ I'm interested in more ways of proving convergence/divergence for this series.

My thoughts:

$$\sum\limits_{n\geq 1}\dfrac{1}{(3n-1)(3n+1)}$$

I'm going to use Integral Test

let $f(n)=\dfrac{1}{(3n-1)(3n+1)}$

we note that $f$ is a continuous, positive and decreasing function on the interval $[1,+\infty ($ then

$$\sum_{n=1}^\infty f(n) \text{ Convergent } \iff \int_1^\infty f(x)\,dx \text{ Convergent }$$

so let's try to compute $$\int_1^\infty f(n)\,dn=\int_1^\infty \dfrac{1}{(3n-1)(3n+1)}\,dn $$

Substitution for $n=\dfrac{1}{3}u$ and $dn=\dfrac{1}{3}du$ we got :

\begin{align} \int_1^\infty f(n)\,dn &=\int_1^\infty \dfrac{1}{(3n-1)(3n+1)}\,dn \\ &=\int_1^\infty \dfrac{1}{(3(\dfrac{1}{3}u)-1)(3(\dfrac{1}{3}u)+1)}\,\dfrac{1}{3}du \\ &=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u-1)(u+1)}\,du \\ &=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u^2-1)}\,du \\ \end{align} note that $\left(arctanh(u)\right)'=\left(\tanh^{-1}(h)\right)'=\dfrac{1}{1-u^{2}}$ so \begin{align} \int_1^\infty f(n)\,dn &=\dfrac{1}{3}\int_1^\infty \dfrac{1}{(u^2-1)}\,du \\ &=-\dfrac{1}{3}\int_1^\infty \dfrac{1}{(1-u^2)}\,du \\ &=-\dfrac{1}{3} \left(\tanh^{-1}(u)\right) \biggl|_{1}^{+\infty} \\ &=-\dfrac{1}{3} \left(\tanh^{-1}(3n)\right) \biggl|_{1}^{+\infty} \\ \end{align}

so can't go ahead beause i don't know the limit of $tanh^{-1}$ i can use woflrame to do that but suppose im in the contest what gonna do so my questions:

  • Is my proof correct
  • since the calculation is not easy in this case I'm interested in more ways of proving convergence/divergence for this series.
Did
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Educ
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    Another way to prove convergence: for $n > 1$ we have the inequalities $$\frac{1}{(3n-1)(3n+1)} < \frac{1}{(3n-1)^2} < \frac{1}{(3(n-1))^2}$$ so we can compare with $$\frac{1}{9}\sum_{n=2}^{\infty}\frac{1}{(n-1)^2}=\frac{1}{9}\sum_{n=1}^{\infty}\frac{1}{n^2}$$ which converges as a $p$-series with $p=2$. –  Sep 22 '15 at 16:41
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    In the OP there is way too much work. And the first way is problematic, though fixable. Use Comparison with $\sum \frac{1}{8n^2}$ or Limit Comparison with $\sum \frac{1}{9n^2}$. – André Nicolas Sep 22 '15 at 16:42
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    The most direct proof is to note that $\sum_{n=1}^m\frac1{(3n-1)(3n+1)} $ = $\sum_{n=1}^m\left(\frac1{6n-2}-\frac1{6n+2}\right) $ $<\sum_{n=1}^m\left(\frac1{6n-3}-\frac1{6(n+1)-3}\right)=\frac13-\frac1{3m+3}$ by telescoping – Hagen von Eitzen Sep 22 '15 at 16:47
  • Or just show $1/(9n^2-1)\le 1/n^2$ for all $n.$ To the OP: You have $n$ in the integrals $dx;$ you should edit those. – zhw. Sep 22 '15 at 17:00
  • @zhw. thank u i fixed – Educ Sep 22 '15 at 17:06

3 Answers3

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There is an easier way to check convergence: $$0\leq \sum_{n\geq 1}\frac{1}{(3n-1)(3n+1)}\leq \sum_{n\geq 1}\frac{1}{8n^2}=\frac{\pi^2}{48}.$$ We may compute the value of the series by considering that: $$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{(3n-1)(3n+1)}&=&\frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}\left(x^{3n-2}-x^{3n}\right)\,dx\\&=&\frac{1}{2}\int_{0}^{1}\frac{1+x}{1+x+x^2}\,dx\\&=&\color{red}{\frac{1}{2}\left(1-\frac{\pi}{3\sqrt{3}}\right)}.\end{eqnarray*}$$

Jack D'Aurizio
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Checking Convergence

Applying a Telescoping Series, we get $$ \begin{align} \sum_{k=1}^N\frac1{(3n-1)(3n+1)} &\le\sum_{k=1}^N\frac1{(3n-2)(3n+1)}\\ &=\frac13\sum_{k=1}^N\left(\frac1{3n-2}-\frac1{3n+1}\right)\\ &=\frac13\left(1-\frac1{3N+1}\right) \end{align} $$ Therefore, $$ \sum_{k=1}^\infty\frac1{(3n-1)(3n+1)}\le\frac13 $$

Evaluating the Sum

Using $(9)$ from this answer, $$ \begin{align} \sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-1}\right) &=\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-1/3}\right)\\ &=\frac13H_{-1/3}\\[3pt] &=-\frac12\log(3)+\frac\pi{6\sqrt3}\tag{1} \end{align} $$ Subtracting $(8)$ from $(7)$ in that same answer gives $$ \begin{align} \sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n+1}\right) &=\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-2}\right)+\overbrace{\sum_{n=1}^\infty\left(\frac1{3n-2}-\frac1{3n+1}\right)}^{\text{telescoping series}}\\ &=\frac13\sum_{n=1}^\infty\left(\frac1n-\frac1{n-2/3}\right)+1\\ &=\frac13H_{-2/3}+1\\[3pt] &=-\frac12\log(3)-\frac\pi{6\sqrt3}+1\tag{2} \end{align} $$ Partial fractions and $(1)$ and $(2)$ yields $$ \begin{align} \sum_{n=1}^\infty\frac1{(3n-1)(3n+1)} &=\frac12\sum_{n=1}^\infty\left(\frac1{3n-1}-\frac1{3n+1}\right)\\ &=\frac12\left[\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n+1}\right) -\sum_{n=1}^\infty\left(\frac1{3n}-\frac1{3n-1}\right)\right]\\[3pt] &=\frac12-\frac\pi{6\sqrt3}\tag{3} \end{align} $$

Community
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robjohn
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The method proposed is one way forward. To carry out the integration, we use partial fraction expansion to write

$$\begin{align} \int_1^{\infty}\frac{1}{(3x-1)(3x+1)}\,dx&=\int_1^{\infty}\left(\frac{1/2}{3x-1}-\frac{1/2}{3x+1}\right)\,dx\\\\ &=\left.\frac12\log\left(\frac{3x-1}{3x+1}\right)\right|_{1}^{\infty}\\\\ &=\frac12 \log 2 \end{align}$$

and conclude that the series converges.

At the request of the OP, we use limit comparison test to show convergence. We recall that

$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

Since $$\lim_{n\to \infty}\frac{1/n^2}{1/(9n^2-1)}=9<\infty$$then the series of interest converges.

Mark Viola
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  • Could you also make it with taylor expanasion and Limit Comparison please – Educ Sep 22 '15 at 16:51
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    @Educ I added a section that uses the limit comparison test. I'm not sure what you are looking for with respect to using Taylor expansion. – Mark Viola Sep 22 '15 at 17:01