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Sum of a power series in terms of x

  • #1
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 ex.

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

Answers and Replies

  • #2
jbunniii
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Do you know this series?

[tex]\sum_{n=0}^{\infty} \frac{x^n}{n!}[/tex]
 
  • #3
Do you know this series?

[tex]\sum_{n=0}^{\infty} \frac{x^n}{n!}[/tex]
yes, thats the expansion for ex. Its the (n+2)! in the original problem thats confusing me.
 
  • #4
dx
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Factor out [tex] \frac{1}{(x + 1)^2} [/tex].
 
  • #5
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]
 
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  • #6
dx
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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 ex+1?
 
  • #7
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]?
 
  • #8
dx
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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]
 
  • #9
[tex]

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

[/tex]

yes?
 
  • #10
dx
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Correct.
 
  • #11
thank you for the help dx :!!)
 

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