Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is given that:

$$v_n = n(n+1)(n+2)\;...\;(n+m)$$ $$and$$ $$u_n = (n+1)(n+2)\;...\;(n+m)$$

$i.$ Verify that:

$$v_{n+1} - v_n = (m+1)(n+1)(n+2)\;...\;(n+m)$$

I started off by inspecting $v_n$:

$$v_1 = 1(2)(3)\;...\;(1+m) = (m+1)!$$ $$v_2 = 2(3)(4)\;...\;(2+m) = \frac{(m+2)!}{1}$$ $$v_3 = 3(4)(5)\;...\;(3+m) = \frac{(m+3)!}{2!}$$ $$\implies v_n = \frac{(m+n)!}{(n-1)!}$$ $$\implies v_{n+1} - v_n = \frac{(m+n+1)!}{n!} - \frac{(m+n)!}{(n-1)!}$$

That's as far as I could get. How can I verify the above equation?

$ii.$ Find $\sum_{n=1}^N u_n$ in terms of $m$ and $N$

Since $u_n$ can be written as $\frac{v_n}{n}$:

$$\implies \sum_{n=1}^N u_n = \sum_{n=1}^N \frac{v_n}{n} = \sum_{n=1}^N \frac{(m+n)!}{n(n-1)!} = \sum_{n=1}^N \frac{(m+n)!}{n!}$$

How can I sum the above equation given the factorial complication?

share|cite|improve this question
up vote 1 down vote accepted

As alex.jordan has aleady pointed out, there’s no need to work with factorials:

$$\begin{align*} v_{n+1}-v_n&=\prod_{k=0}^m(n+1+k)-\prod_{k=0}^m(n+k)\\\\ &\overset{(1)}=\prod_{k=1}^{m+1}(n+k)-\prod_{k=0}^m(n+k)\\\\ &\overset{(2)}=(n+m+1)\prod_{k=1}^m(n+k)-n\prod_{k=1}^m(n+k)\\\\ &\overset{(3)}=\Big((n+m+1)-n\Big)\prod_{k=1}^m(n+k)\\\\ &=(m+1)\prod_{k=1}^m(n+k)\;, \end{align*}$$

as desired. And $\prod_{k=1}^m(n+k)=u_n$, so we’ve established that $u_n=\frac1{m+1}(v_{n+1}-v_n)$. Thus,

$$\begin{align*} \sum_{n=1}^Nu_n&=\frac1{m+1}\sum_{n=1}^N(v_{n+1}-v_n)\\\\ &=\frac1{m+1}\left(\sum_{n=1}^Nv_{n+1}-\sum_{n=1}^Nv_n\right)\\\\ &\overset{(1)}=\frac1{m+1}\left(\sum_{n=2}^{N+1}v_n-\sum_{n=1}^Nv_n\right)\\\\ &\overset{(2)}=\frac1{m+1}\left(v_{N+1}+\sum_{n=2}^Nv_n-\sum_{n=2}^Nv_n-v_1\right)\\\\ &\overset{(3)}=\frac1{m+1}(v_{N+1}-v_1)\;. \end{align*}$$

In the steps marked $(1)$ I’ve shifted the index of the first product (or sum) by $1$ to give the general expressions inside the two products (or sums) identical forms. In the steps marked $(2)$ I’ve separated the extra factor (or term) from each of the two products (or sums), so that in what’s left the index runs over the same range of values. And in the steps marked $(3)$ I’ve combined the identical products (or sums). This is a fairly common pattern of manipulation.

It’s a good idea to learn perform manipulations like these on products and summations, but of course you can always write them out longhand with ellipses for the missing terms, as alex.jordan did in his first comment (with the blue terms).

share|cite|improve this answer
It makes a lot more sense to me, to solve the problem like this. This question was worth 5 marks in FP1, which gives me over 13 minutes to solve it. Like this it should take around 5 minutes. Thanks a lot. – Ozzy Mar 13 '13 at 6:47
@Ozzy: You’re very welcome. – Brian M. Scott Mar 13 '13 at 8:05
I should correct myself... FP1 is 3hr/100marks so 9 minutes for this question. Still great though – Ozzy Mar 13 '13 at 8:56

For $i$, factor out $(n+1)\cdots(n+m)$ from $v_{n+1}-v_n$.

For $ii$, $\sum_{n=0}^Nu_n=\sum_{n=0}^N\frac{v_{n+1}-v_n}{m+1}$, using the results of $i$. Pull out the $m+1$ and sum with the telescoping series method, yielding $\frac{1}{m+1}(v_{N+1}-v_0)$, which equals $\frac{(m+N+1)!}{(m+1)N!}$.

share|cite|improve this answer
Ohh I get it so for $i.$ I expand the first factor of $(m+n+1)!$ to get $(m+n+1)(m+n)!$ and then factor out the $(m+n)!$ from both sides. That's a great help, thanks! – Ozzy Mar 12 '13 at 7:09
Yes although I was looking straight at the definition of $v_n$, not at the formula that you have found for it. $v_{n+1}-v_n={\color{blue}{(n+1)\cdots(n+m)}}(n+m+1)-n{\color{blue}{(n+1)\cdots(‌​n+m)}}=(m+1){\color{blue}{(n+1)\cdots(n+m)}}$ – alex.jordan Mar 12 '13 at 7:14
I've got to this point: $v_{n+1}-v_n = \frac{(m+n+1)(m+n)! - n(m+n)!}{n!} = \frac{(m+n)!}{n!} \bigg( m+n+1-n(m+n)! \bigg)$ In your edited comment I can see how the terms in blue cancel, but what about the denominator of $n!\;$? – Ozzy Mar 12 '13 at 7:24
In your last comment, the last factor in your equation is not correct. After factoring out $(m+n)!$ your last factor should be $\bigg(m+n+1-n\bigg)$, or just $\bigg(m+1\bigg)$. That yields $\frac{(m+n)!}{n!}(m+1)$, which equals $(m+1)(n+1)\cdots(n+m)$ when you expand the factorials. And that's the definition of $u_n$. – alex.jordan Mar 12 '13 at 17:20
$m+n$ is greater than $n$. So if you really keep expanding that numerator you will eventually reach $(n)(n-1)(n-2)\cdots$ in the numerator too. And these factors will cancel with the entire denominator. What remains in the numerator will be the factors right up to $(n)$, the last of which was $(n+1)$. So what remains after cancellation is $(m+n)(m+n-1)\cdots(n+1)$. – alex.jordan Mar 13 '13 at 6:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.