Power series - finding the sum

In summary, you have found the sum of a geometric series with a=x2 and r=xn. The equation for the sum is: f(x)=xe^{x^3}. You can find the sum of a geometric series with a=x2 and r=xn by Differentiating and using the identities for geometric series.
  • #1
nonaa
17
0
[tex]\sum_{n=0}^{\infty}(n+1)(n+2)x^n[/tex]
 
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  • #2
Notice that (n+2)(n+1)xn is the second derivative of xn+2

[tex]\sum_{n=0}^\infty x^{n+2}[/tex]

is a geometric series of the form [itex]\sum ar^n[/itex] with a= x2 and r= xn. Use the usual formula for sum of a geometric series to write that in "closed form" and differentiate twice.
 
  • #3
Thank you very much (sun)
 
  • #4
Would you tell me if my solution to the problem is correct?
Again, we are searching for the sum.
[tex]f(x)=\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}=?[/tex]

[tex]\int\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}dx=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{n!}=x.\sum_{n=0}^{\infty}\frac{(x^{3})^n}{n!}=xe^{x^3}[/tex]

[tex]f(x)=(xe^{x^3})'+C=e^{x^3}+3x^3e^{x^3}+C[/tex]

[tex]f(0)=e^{0}+0+C=\sum_{n=0}^{\infty}\frac{(3n+1)0^{3n}}{n!}[/tex]

[tex]1+C = 0 \rightarrow C=-1[/tex]
 
  • #5
nonaa said:
Would you tell me if my solution to the problem is correct?
Again, we are searching for the sum.
[tex]f(x)=\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}=?[/tex]

[tex]\int\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}dx=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{n!}=x.\sum_{n=0}^{\infty}\frac{(x^{3})^n}{n!}=xe^{x^3}[/tex]

[tex]f(x)=(xe^{x^3})'+C=e^{x^3}+3x^3e^{x^3}+C[/tex]

[tex]f(0)=e^{0}+0+C=\sum_{n=0}^{\infty}\frac{(3n+1)0^{3n}}{n!}[/tex]

[tex]1+C = 0 \rightarrow C=-1[/tex]

Almost. Why do you introduce the constant C? You have shown:
[tex]\int f(x) \text{ d}x = xe^{x^3}[/tex]
so you get,
[tex]f(x) = \left(xe^{x^3}\right)'[/tex]
It's only when integrating you add the constant. Also you derive the wrong value for C because in the sum:
[tex]\sum_{n=0}^\infty \frac{{x}^{3n}}{n!}[/tex]
the first term actually doesn't make any sense for x=0 since it has 0^0, but we always interpret the first term to be 1 (this is just shorthand when dealing with Taylor series) so:
[tex]\sum_{n=0}^\infty \frac{{0}^{3n}}{n!} = 1 + 0^3/1! + 0^6/2! + \cdots = 1[/tex]
Thus C=0 and we get rid of the constant which shouldn't have been there in the first place.
 
  • #6
Thank you :)
 
  • #7
Ok, this is the last one, I promise :blushing:

[tex]\sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}=?[/tex]

It's a little confusing for me because there is no x. I tried with [tex]1^n = x^n[/tex] and finding the sum
[tex]\sum_{n=0}^{\infty}\frac{(-x)^n}{2n+1}=?[/tex]
but no result...
 
  • #8
Try working with (we use 2n+1 for exponent to cancel the denominator and get a geometric series when differentiating):
[tex]f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}[/tex]
Differentiate it, use the identities for geometric series and integrate again. It turns out f is actually a function you probably know very well.
 
  • #9
nonaa said:
Ok, this is the last one, I promise :blushing:

[tex]\sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}=?[/tex]

It's a little confusing for me because there is no x. I tried with [tex]1^n = x^n[/tex] and finding the sum
[tex]\sum_{n=0}^{\infty}\frac{(-x)^n}{2n+1}=?[/tex]
but no result...

rasmhop said:
Try working with (we use 2n+1 for exponent to cancel the denominator and get a geometric series when differentiating):
[tex]f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}[/tex]
Differentiate it, use the identities for geometric series and integrate again. It turns out f is actually a function you probably know very well.
And, of course, after you have found it as a function of x, set x= 1 to get your numerical sum.
 

1. What is a power series?

A power series is a representation of a function as an infinite sum of terms. It is typically written in the form of f(x) = a0 + a1x + a2x2 + a3x3 + ... + anxn + ...

2. How do you find the sum of a power series?

To find the sum of a power series, you need to use a method called "term-by-term integration." This involves integrating each term of the series individually and then adding them together. The result will be a new power series with a different constant term.

3. What is the radius of convergence for a power series?

The radius of convergence is a value R that represents the distance from the center of a power series to the nearest point where the series converges. It is typically found by using the ratio test, which involves taking the limit of the ratio of consecutive terms in the series.

4. Can a power series converge for some values of x and diverge for others?

Yes, a power series can have a different radius of convergence for different values of x. This means that the series may converge for some values of x and diverge for others. It is important to check the radius of convergence before assuming that a power series will converge for all values of x.

5. What is the Maclaurin series?

The Maclaurin series is a special case of a power series, where the center of the series is at x = 0. It is written in the form of f(x) = a0 + a1x + a2x2 + a3x3 + ... + anxn + ... and is often used to approximate functions near x = 0.

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