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I am trying to find the sum of the above series.

The sum till n terms can be found using power series expansion. However, I'm trying to solve this using the method of difference (a.k.a. Telescoping sum or $V_n$ method).

In this method, the general term is expressed as the difference between two consecutive values of some function. Like the following:

$T_n = V_n - V_{n-1}$

and then the sum is taken which comes to be

$S_n = V_n - V_0$

The general term of the series in question can be represented as a product:

$T_n = n(n+1)^2(n+2)$

But I am unable to represent this as a difference. How can I proceed from here to find the sum?

ab123
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user638500
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  • Duplicate of https://math.stackexchange.com/questions/1903470/sum-of-the-series-1-cdot3-cdot222-cdot4-cdot323-cdot5-cdot42-cdots?rq=1 – R. J. Mathar Jun 22 '21 at 16:14

1 Answers1

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Since $V_n = S_n + V_0$ you have that $V_n - V_{n-1} = S_n - S_{n-1}$ so your search for $V_n$ is identical to finding $S_n$, besides a free constant.

So you may well set $V_n$ to the (known or guessed) sum $S_n$ and write $$V_n = \frac{1}{10} n (n + 1) (n + 2) (n + 3) (2 n + 3)$$

Andreas
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  • I didn't understand what you mean by setting Vn to the sum. I don't know the sum and can't guess it. Also, how did you arrive at the last equation? Can you please elaborate more? – user638500 Jan 27 '19 at 15:25
  • I got $V_n = S_n$ by using power series expansion. As $T_n$ is of order $n^4$, $S_n$ has to be of order $n^5$. You may therefore also set a general formula $V_n = a_0 + a_1 n + a_2 n^2 + a_3 n^3 + a_4 n^4 + a_5 n^5$ and determine the coefficients $a_i$ by setting all powers $n^k$ equal in $V_n - V_{n-1} = T_n$. – Andreas Jan 27 '19 at 19:46
  • I have mentioned in the question that I don't want to use Power series expansion. I am in high school and don't know calculus or Power series expansion. However, thanks for the answer. – user638500 Jan 28 '19 at 03:12
  • I've laid out another path to arrive at $V_n$ - with unknown coefficients to be determined - in my previous comment. Kindly use that one. – Andreas Jan 28 '19 at 08:15