2 questions on Maclaurin series

• Jun 6th 2013, 08:46 AM
jones123
2 questions on Maclaurin series
Hello! I'm stuck on these questions... Can someone help? (Rofl)

First: how would you write f(x) = x / (1 +
x3) as a series?

I mean like

Σ = .....
k=0

I know that 1/(1-x) =

Σ
xk
k=0

But how do i convert this into f(x)?

Second: h
ow would you find the radius of convergence of the following binomial series:

Σ (3k k)
x2k+1
k=0

with (3k k) is like (n k) = n! / k!(n-k)!

I know how to do it when there's
xk in the sommation but with x2k+1?

Thanks for the help, I've been spending hours on these two questions!
(Wondering)
• Jun 6th 2013, 09:01 AM
Prove It
Re: 2 questions on Maclaurin series
Quote:

Originally Posted by jones123
Hello! I'm stuck on these questions... Can someone help? (Rofl)

First: how would you write f(x) = x / (1 +
x3) as a series?

I mean like

Σ = .....
k=0

I know that 1/(1-x) =

Σ
xk
k=0

But how do i convert this into f(x)?

Second: h
ow would you find the radius of convergence of the following binomial series:

Σ (3k k)
x2k+1
k=0

with (3k k) is like (n k) = n! / k!(n-k)!

I know how to do it when there's
xk in the sommation but with x2k+1?

Thanks for the help, I've been spending hours on these two questions!
(Wondering)

1. You could write $\displaystyle \frac{x}{1 + x^3} = x \left[ \frac{1}{1 - \left( -x^3 \right) } \right]$, which you can then write as a geometric series.

2. Use the Ratio Test.
• Jun 6th 2013, 10:05 AM
jones123
Re: 2 questions on Maclaurin series
Hmm I don't quite understand the second one. What difference makes the x^k in stead of x^(2k+1) in the ratio test ?

The radius of convergence of the sum with x^k should be 4/27, the sum of x^(2k+1): 2/sqrt(27)
• Jun 6th 2013, 02:41 PM
HallsofIvy
Re: 2 questions on Maclaurin series
You have $\sum_{k= 0}^\infty\begin{pmatrix}3k \\ k\end{pmatrix}x^{2k+1}= \tex]\sum_{k= 0}^\infty \frac{(2k+ 1)!}{k! (k+1)!} x^{2k+1}$
To use the "ratio test" to determine whether the series $\sum_{k=0}^\infty} a_n$ converges, look at $\lim_{k\to\infty}\frac{a_{k+1}}{a_k}$

Here, $a_k= \frac{(2k+1)!}{k!(k+1)!}x^{2k+1}$ so $a_{k+1}= \frac{(2(k+1)+1)!}{(k+1)!((k+1)+1)!}x^{2(k+1)+1}= \frac{(2k+3)!}{(k+1)!(k+ 2)!}x^{2k+ 3}$

and so $\frac{a_{k+1}}{a_k}= \frac{(2k+3)!}{(k+1)!(k+2)!}\frac{k!(k+1)!}{(2k+1) !}\frac{x^{2k+3}}{x^{2k+1}}$ $= \frac{(2k+ 3)!}{(2k+1)!}\frac{k!}{(k+ 1)!}\frac{(k+1)!}{(k+2)!}x^2$

Can you complete it from there?