0

$$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$

Does calculating the limit ($\frac 1 4$) suffice for showing that it's convergent?

If not, how could I show it?

  • Quotient criterion - the ratio of two consecutive sequence terms is $1$, so it won't work.
  • Can it be done by comparing with the harmonic series? I don't see how I can transform the fraction?
Starlight
  • 1,680
Math for fun
  • 590
  • You have found the value of the sum, which is finite, thus it converges. However, you can directly say that it converges without computing the sum by comparing its terms with an appropriate series. For instance, each of the terms in your series is less than $1/n^3$. – Alberto Debernardi Nov 10 '17 at 10:03
  • https://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2 – lab bhattacharjee Nov 10 '17 at 10:04
  • 1
    @AlbertoDebernardi Finding a finite sum is equivalent to convergence only because the series has only positive terms, beware not to induce hasty generalization for OP. – zwim Nov 10 '17 at 10:06
  • The harmonic series cannot help you, as it is divergent. But how did you obtain that the sum is $1/4$ ?? –  Nov 10 '17 at 10:09

3 Answers3

0

Hint the general term of this serie is inferior to $1/n^3$, apply the comparisaon test.

Tsemo Aristide
  • 87,475
0

Normally calculating the value of the series is enough, if you did not do a mistake. But if you are only asked to show convergence then it is easier to find an upper bound.

$$\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)}< \sum_{n=1}^{\infty}\dfrac{1}{n(n+0)2}= \dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{1}{2}\dfrac{\pi^2}{6}$$

As the sum is monotonically adding positive terms and as we found an upper bound it is clear that the series does converge to a limit.

MrYouMath
  • 15,833
0

$S_n:= \sum_{k=1}^{n} \dfrac{1}{n(n+1)(n+2)} \lt$

$\sum_{k=1}^{n}\dfrac{1}{n(n+1)} =$

$\sum_{k=1}^{n}(\dfrac{1}{n} -\dfrac{1}{n+1}) =$

$=1- \dfrac{1}{n+1} \lt 1.$

$S_n$ is strictly increasing and bounded above.

$ \Rightarrow:$

$S_n$ is convergent.

Peter Szilas
  • 20,344
  • 2
  • 17
  • 28