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As I calculated the sum of the serie above doesn't exist(sum doesn't converge). How can I prove it using the double computing(combinatorical method)?

Zauberkerl
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    Are you asking for an explicit formula? –  Feb 10 '16 at 20:10
  • yes, but I think it doesn't exist – Zauberkerl Feb 10 '16 at 20:10
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    Certainly, the limit as $n$ approaches infinity will not converge (how could it? the limit of the summands does not approach zero). The question title implies that you are wishing to find a closed form for the partial sums, while the body of the question seems to imply that you wish to prove that the series doesn't converge. – JMoravitz Feb 10 '16 at 20:12
  • The example sounds exactly so: Find a closed formula for the next sum... I've checked that this serie is convergent, so I thought that a closed formula doesn't exist. – Zauberkerl Feb 10 '16 at 20:14
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    In general, you can use Faulhaber's formula upon expanding the summand. – user170231 Feb 10 '16 at 20:21
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    All sums of the form

    $$\sum_{k=1}^nP(k),$$ were $P$ is a polynomial of degree $d$ can be expressed in a closed-form, which is a polynomial in $n$ of the degree $d+1$.

    This can be established using the Faulhaber formula, or more directly.

    Let $$Q(n):=\sum_{k=1}^nP(k).$$

    Then

    $$Q(n)-Q(n-1)=P(n).$$

    By identifying the coefficients, you obtain a linear system of $d+1$ equations in $d+1$ unkowns $q_1,q_2,\cdots q_{d+1}$ (indeed, $Q(0)=q_0=0$).

     –  Feb 10 '16 at 20:33

1 Answers1

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use the hockey stick identity.

What we want is $\sum_{i=3}^{n}\binom{n}{3}6$

It is a known fact $\sum_{i=k}^n\binom{i}{k}=\binom{n+1}{k+1}$

Hence $\sum_{i=3}^{n}i(i-1)(i-2)=6\binom{n+1}{4}$

Asinomás
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