Sum of a power series in terms of x

[OrbsAndSuch]

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

The hint i was given was compare it to the Taylor series expansion of e^x.

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

 

[jbunniii]

Do you know this series?

[tex]\sum_{n=0}^{\infty} \frac{x^n}{n!}[/tex]

 

[OrbsAndSuch]

Do you know this series?

[tex]\sum_{n=0}^{\infty} \frac{x^n}{n!}[/tex]

yes, that's the expansion for e^x. Its the (n+2)! in the original problem that's confusing me.

 

[dx]

Factor out [tex] \frac{1}{(x + 1)^2} [/tex].

 

[OrbsAndSuch]

ok i factored and came up with:

[tex]

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

[/tex]

Do the indices matter? If not then it becomes

[tex]\frac{e^{x+1}}{(x+1)^2}[/tex]

 

[dx]

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

[tex] \sum_2^\infty \frac{(x+1)^n}{n!} [/tex]

to make it equal to e^x+1?

 

[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)

[tex]

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

[/tex]?

 

[dx]

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

[tex] 1 + (1 + x) + \sum_2^\infty \frac{(x+1)^n}{n!} = e^{x+1} [/tex]

This means

[tex] \sum_2^\infty \frac{(x+1)^n}{n!} = e^{x+1} - 2 - x [/tex]

Substitute this back in

[tex] \frac{1}{(x+1)^2} \sum_2^\infty \frac{(x+1)^n}{n!} [/tex]

 

[OrbsAndSuch]

[tex]

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

[/tex]

yes?

 

[dx]

Correct.

 

[OrbsAndSuch]

thank you for the help dx :!)