# Seperating a Summation problem.

1. May 3, 2008

### DKATyler

[SOLVED] Seperating a Summation problem.

1. The problem statement, all variables and given/known data
The Problem:
Separate a sum into 2 pieces (part of a proof problem).

Using: $$X= \sum^{n}_{k=1}\frac{n!}{(n-k)!}$$

Solve in relation to n and X:
$$\sum^{n+1}_{k=1}\frac{(n+1)!}{(n+1-k)!}$$

2. Relevant equations
???

3. The attempt at a solution
$$\sum^{n}_{k=1}[\frac{(n+1)!}{(n+1-k)!}]+\frac{(n+1)!}{(n+1-[n+1])!}$$

$$\sum^{n}_{k=1}[\frac{(n)!}{(n-k)!}*\frac{(n+1)}{(n+1-k)}]+\frac{(n+1)!}{(n+1-[n+1])!}$$

$$(n+1)*\sum^{n}_{k=1}[\frac{(n)!}{(n-k)!}*\frac{1}{(n+1-k)}]+(n+1)!}$$

I think this is fairly close but, I have no way of getting rid of the 1/(n+1-k) term.

2. May 4, 2008

### Defennder

Can you show that $$\sum_{k=2}^n \frac{n!}{(n-k+1)!} \ + \ \frac{n!}{(n+1-(n+1))!} = \sum^{n}_{k=1}\frac{n!}{(n-k)!}$$?

If you do that, you can express $$\sum^{n+1}_{k=1}\frac{(n+1)!}{(n+1-k)!}$$ as a summation starting from k=2. Then you should be able to get the desired expression.

Try it out for some values of k and n, then you'll see a pattern.

In general the pattern is $$\sum_{k=1}^{n}f(k) = \sum_{k=2}^{n} f(k-1)\ +\ f(n)$$

3. May 4, 2008

### DKATyler

That helps quite a bit.

$$n!+\sum^{n}_{k=2}\frac{n!}{(n+1-k)!}=\sum^{n}_{k=1}\frac{n!}{(n-k)!}=X$$

So continuing from this step:
$$(n+1)*\sum^{n}_{k=1}\frac{n!}{(n-k+1)!}+\frac{(n+1)!}{(n+1-[n+1])!}$$

changing the Index and adding/subtracting n!
$$(n+1)*(1-n!+n!+\sum^{n}_{k=2}\frac{n!}{(n-k+1)!})+(n+1)!$$

Solves the Equation in terms of n and X:
$$(n+1)*(1-n!+X)+(n+1)!$$

Yep, that worked, now I can complete the rest of the proof :) Thank you very much. How to mark this "[solved]?"

4. May 4, 2008

### Defennder

Go to your first post in this thread, at the top right hand corner of the post marked "Thread Tools"