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I have to solve this series transforming it into a telescopic sequence $$\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$$

But I'm lost in the calculation!

Cornman
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Anne
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    First compute the partial fraction decomposition of the rational function being summed. – Dave Jan 02 '18 at 18:24
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    See also : https://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2 – lab bhattacharjee Jan 02 '18 at 18:38
  • "But I'm lost in the calculation!" If this is true, you have some "calculation" to show, please do that. – Did Jan 03 '18 at 07:04
  • Using Approach0 you can find the following question: https://math.stackexchange.com/questions/1541224/find-sum-of-infinite-anharmonic-series – Martin Sleziak Jan 03 '18 at 12:52
  • Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Jan 03 '18 at 12:53

3 Answers3

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There is a general solution for this type of problem: $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}.$$

Observe that

$$\sum_{k\geq 1} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)}=\sum_{k\geq 1} \frac{(k-1)!}{(k+N)!}=\frac{1}{N!}\sum_{k\geq 1}\frac{\Gamma(k)\Gamma(N+1)}{\Gamma(k+N+1)}=\frac{1}{N!}\sum_{k\geq 1} B(k,N+1).$$

Consequently, \begin{align} \sum_{k=1}^{\infty} \frac{1}{k(k+1)\cdot\dotso\cdot (k+N)} & = \frac{1}{N!}\sum_{k\geq 1}\int_0^1 x^{k-1}(1-x)^{N}\,dx \\ & =\frac{1}{N!}\int_0^1\left(\sum_{k\geq 1}x^{k-1}\right)(1-x)^N\, dx \\ &=\frac{1}{N!}\int_0^1 (1-x)^{N-1}\, dx \\ & =\frac{1}{N\cdot N!}. \end{align}

Here we have $N=3$. Hence, the answer is $\frac{1}{3\cdot 3!}=\frac{1}{18}$.

Math Lover
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Cornman
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Hint:

Look at $$\frac{1}{k(k+1)(k+2)} - \frac{1}{(k+1)(k+2)(k+3)}$$

It is not exactly the general term of your sum, but you only need to divide by a constant such that it is.

Math Lover
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You can also go the hard way: write $$ \frac{1}{x(x+1)(x+2)(x+3)}= \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+\frac{D}{x+3} $$ so $$ A(x+1)(x+2)(x+3)+Bx(x+2)(x+3)+Cx(x+1)(x+3)+Dx(x+1)(x+2)=1 $$ Now,

  1. for $x=0$: $6A=1$
  2. for $x=-1$: $-2B=1$
  3. for $x=-2$: $2C=1$
  4. for $x=-3$: $-6D=1$

Thus $$ \frac{1}{k(k+1)(k+2)(k+3)}= \frac{1}{6}\left(\frac{1}{k}-\frac{3}{k+1}+\frac{3}{k+2}-\frac{1}{k+3}\right) $$ Hence your summation is $1/6$ of \begin{align} \sum_{k=1}^\infty \frac{1}{k} -3\sum_{k=1}^\infty \frac{1}{k+1} +3\sum_{k=1}^{\infty} \frac{1}{k+2} -\sum_{k=1}^{\infty} \frac{1}{k+3} & =\sum_{k=1}^\infty \frac{1}{k} -3\sum_{k=2}^\infty \frac{1}{k} +3\sum_{k=3}^{\infty} \frac{1}{k} -\sum_{k=4}^{\infty} \frac{1}{k} \\[6px] &= \left(1+\frac{1}{2}+\frac{1}{3}\right)-3\left(\frac{1}{2}+\frac{1}{3}\right) +3\frac{1}{3} \\[6px] &=\frac{1}{3} \end{align} Therefore the sum of your series is $$ \frac{1}{6}\cdot\frac{1}{3}=\frac{1}{18} $$

egreg
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