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 \Sigma^{\infty}_{n=0}\frac{(x+1)^n}{(n+2)!}

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?

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

 

[OrbsAndSuch]

Do you know this series?

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

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

 

[dx]

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

 

[OrbsAndSuch]

ok i factored and came up with:

<br /> <br /> \frac{1}{(x+1)^2} \sum_2^\infty \frac{(x+1)^n}{n!} <br /> <br />

Do the indices matter? If not then it becomes

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

 

[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 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)

<br /> <br /> 1 + (x+1) + \frac{e^{x+1}}{(x+1)^2}<br /> <br />?

 

[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!}

 

[OrbsAndSuch]

<br /> <br /> \frac{e^{x+1} - x - 2}{(x+1)^2}<br /> <br />

yes?

 

[dx]

Correct.

 

[OrbsAndSuch]

thank you for the help dx :!)