How to Find the Sum of Power Series Without a Variable?

[nonaa]

$$\sum_{n=0}^{\infty}(n+1)(n+2)x^n$$

 

[HallsofIvy]

Notice that (n+2)(n+1)xⁿ is the second derivative of xⁿ⁺²

$$\sum_{n=0}^\infty x^{n+2}$$

is a geometric series of the form $\sum ar^n$ 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.

 

[nonaa]

Thank you very much (sun)

 

[nonaa]

Would you tell me if my solution to the problem is correct?
Again, we are searching for the sum.
$$f(x)=\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}=?$$

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

$$f(x)=(xe^{x^3})'+C=e^{x^3}+3x^3e^{x^3}+C$$

$$f(0)=e^{0}+0+C=\sum_{n=0}^{\infty}\frac{(3n+1)0^{3n}}{n!}$$

$$1+C = 0 \rightarrow C=-1$$

 

[rasmhop]

Would you tell me if my solution to the problem is correct?
Again, we are searching for the sum.
$$f(x)=\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}=?$$

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

$$f(x)=(xe^{x^3})'+C=e^{x^3}+3x^3e^{x^3}+C$$

$$f(0)=e^{0}+0+C=\sum_{n=0}^{\infty}\frac{(3n+1)0^{3n}}{n!}$$

$$1+C = 0 \rightarrow C=-1$$

Almost. Why do you introduce the constant C? You have shown:
$$\int f(x) \text{ d}x = xe^{x^3}$$
so you get,
$$f(x) = \left(xe^{x^3}\right)'$$
It's only when integrating you add the constant. Also you derive the wrong value for C because in the sum:
$$\sum_{n=0}^\infty \frac{{x}^{3n}}{n!}$$
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:
$$\sum_{n=0}^\infty \frac{{0}^{3n}}{n!} = 1 + 0^3/1! + 0^6/2! + \cdots = 1$$
Thus C=0 and we get rid of the constant which shouldn't have been there in the first place.

 

[nonaa]

Thank you :)

 

[nonaa]

Ok, this is the last one, I promise

$$\sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}=?$$

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

 

[rasmhop]

Try working with (we use 2n+1 for exponent to cancel the denominator and get a geometric series when differentiating):
$$f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$$
Differentiate it, use the identities for geometric series and integrate again. It turns out f is actually a function you probably know very well.

 

[HallsofIvy]

Ok, this is the last one, I promise

$$\sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}=?$$

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

Try working with (we use 2n+1 for exponent to cancel the denominator and get a geometric series when differentiating):
$$f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$$
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.