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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?

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  • $\begingroup$ 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. $\endgroup$ Mar 25, 2012 at 14:46
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    $\begingroup$ 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, 2012 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. $\endgroup$ Mar 25, 2012 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, 2013 at 15:11

2 Answers 2

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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*}

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  • $\begingroup$ 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, 2014 at 5:41
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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.

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