On my last exam there was the question if the series $\sum_{n=2}^{\infty}\frac{1}{(n-1)n(n+1)}$ converges and which limit it has. During the exam and until now, I am not able to solve it. I tried partial fraction decomposition, telescoping sum, etc. But I am not able to find the partial sum formula (Wolfram|Alpha):

$$ \sum_{n=2}^{m}\frac{1}{(n-1)n(n+1)} = \frac{m^2+m-2}{4m(m+1)}. $$

Could somebody push me in the right direction? Is there any trick or scheme how to find partial sum formulas for given series?

  • $\begingroup$ Each term is less than $1/n^2$, so it converges. You don't need to know what a series converges to to know that it converges. $\endgroup$ – Thomas Andrews Mar 25 '12 at 14:46
  • 1
    $\begingroup$ Two things to observe(Unrelated to the Math): Please do not sign your posts with signature as faq explicitly lists it! Accepting answers is a sign of appreciation for someone who put in effort compiling an answer for you. Please accept answers which you think helped you a lot in solving that problem or cleared up your concepts and whatever. It is done by clicking on the tick mark besides every answer. $\endgroup$ – user21436 Mar 25 '12 at 14:51
  • $\begingroup$ The ratio test with $1/n^3$ proves it converges. As often happens, finding the sum takes more work than that. $\endgroup$ – Michael Hardy Mar 25 '12 at 16:29
  • $\begingroup$ martini's answer is the right thing to do. Just wanted to add that if you know the answer (given by wolfram) you can just prove the it with recursion arguments $\endgroup$ – Thomas Dec 15 '13 at 15:11

So let's try partial fraction decomposition. Writing $$ \frac 1{(n-1)n(n+1)} = \frac a{n-1} + \frac bn + \frac c{n+1} $$ we obtain $$ 1 = a(n^2 + n) + b(n^2 - 1) + c(n^2 - n) $$ and therefore \begin{align*} 1 &= -b\\ 0 &= a - c\\ 0 &= a + b + c. \end{align*} This gives $b = -1$, $a = c = \frac 12$. Hence \begin{align*} \sum_{n=2}^m \frac 1{(n-1)n(n+1)} &= \sum_{n =2}^m \frac 1{2(n-1)} - \sum_{n=2}^m \frac 1n + \sum_{n=2}^m \frac 1{2(n+1)}\\ &= \frac 12 + \sum_{n=2}^{m-1} \frac 1{2n} - \sum_{n=2}^m \frac 1n + \sum_{n=3}^m \frac 1{2n} + \frac 1{2(m+1)}\\ &= \frac 12 + \frac 14 - \frac 12 - \frac 1m + \frac 1{2m} + \frac 1{2m+2}\\ &= \frac 14 + \frac{-2(m+1) + m+1 + m}{2m(m+1)}\\ &= \frac 14 + \frac{-1}{2m(m+1)}\\ &= \frac{m(m+1) - 2}{4m(m+1)}. \end{align*}

  • $\begingroup$ I got little confuse here, why the term $\frac{1}{4}$ doesnt vanish? In my calculation the third step from below, I got $$=\frac{1}{2}+ (\frac{1}{4}+ \frac{1}{6}+\dots+ \frac{1}{2(m-1)})-( \frac{1}{2}+ \frac{1}{3}+\dots+ \frac{1}{m})+( \frac{1}{6}+ \frac{1}{8}+\dots+ \frac{1}{2m})+ \frac{1}{2(m+1)} = \\ =-( \frac{1}{3}+ \frac{1}{5}+\dots+ \frac{1}{m})+(\frac{1}{6}+\frac{1}{8}+\dots+ \frac{1}{2m})+\frac{1}{2(m+1)}$$ How can I get to the next step? $\endgroup$ – DadangAH Feb 20 '14 at 5:41

What did you try for a telescoping sum? Your denominator here is the product of three successive terms (this is often called a rising or falling factorial, depending on which side you take as your baseline); this points to looking at a difference of terms that are of the same form but with denominators one degree less. In particular, looking at $t_n=\dfrac{1}{n(n+1)}$ then $t_n-t_{n-1}$ $=\dfrac{1}{n(n+1)}-\dfrac{1}{(n-1)n}$ $=\dfrac1n\left(\dfrac1{n+1}-\dfrac1{n-1}\right)$ $=\dfrac1n\left(\dfrac{(n-1)-(n+1)}{(n-1)(n+1)}\right)$ $=\dfrac{-2}{(n-1)n(n+1)}$; in other words, $\dfrac{1}{(n-1)n(n+1)} = -\dfrac12(t_n-t_{n-1})$, and from here the telescopy should be fairly clear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.