Sum of a power series in terms of x

1. May 13, 2009

OrbsAndSuch

Hello, I need to find the sum(as a function of x) of the power series $$\Sigma^{\infty}_{n=0}\frac{(x+1)^n}{(n+2)!}$$

The hint i was given was compare it to the Taylor series expansion of ex.

Im not sure even how to start this problem and any help is much appreciated.

Last edited: May 13, 2009
2. May 13, 2009

jbunniii

Do you know this series?

$$\sum_{n=0}^{\infty} \frac{x^n}{n!}$$

3. May 13, 2009

OrbsAndSuch

yes, thats the expansion for ex. Its the (n+2)! in the original problem thats confusing me.

4. May 13, 2009

dx

Factor out $$\frac{1}{(x + 1)^2}$$.

5. May 13, 2009

OrbsAndSuch

ok i factored and came up with:

$$\frac{1}{(x+1)^2} \sum_2^\infty \frac{(x+1)^n}{n!}$$

Do the indices matter? If not then it becomes

$$\frac{e^{x+1}}{(x+1)^2}$$

Last edited: May 13, 2009
6. May 13, 2009

dx

The indices (you should call them limits) do matter. What two terms do you have to add to

$$\sum_2^\infty \frac{(x+1)^n}{n!}$$

to make it equal to ex+1?

7. May 13, 2009

OrbsAndSuch

I would have to add the first two terms of the sum if the limits started at 0 which would be

1 + (x+1)

$$1 + (x+1) + \frac{e^{x+1}}{(x+1)^2}$$?

8. May 13, 2009

dx

Yes, the first two terms. But your answer is not correct.

$$1 + (1 + x) + \sum_2^\infty \frac{(x+1)^n}{n!} = e^{x+1}$$

This means

$$\sum_2^\infty \frac{(x+1)^n}{n!} = e^{x+1} - 2 - x$$

Substitute this back in

$$\frac{1}{(x+1)^2} \sum_2^\infty \frac{(x+1)^n}{n!}$$

9. May 13, 2009

OrbsAndSuch

$$\frac{e^{x+1} - x - 2}{(x+1)^2}$$

yes?

10. May 13, 2009

dx

Correct.

11. May 13, 2009

OrbsAndSuch

thank you for the help dx :!!)