# Sum of a power series in terms of x

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:

jbunniii
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Do you know this series?

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

Do you know this series?

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

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

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

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

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

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dx
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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?

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

1 + (x+1)

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

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

This means

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

Substitute this back in

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

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

yes?

dx
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Correct.

thank you for the help dx :!!)