Question

If f(^(n))(0) = (n+1)! for 0, 1, 2, ..., find the Maclaurin series for f.

Find its radius of convergence R.

Answer 1

The Maclaurin series for a function \(f\) is
\[\sum_{n=0}^\infty\frac{1}{n!}f^{(n)}(0)x^n.\]
Since you're given that \(f^{(n)}(0)=(n+1)!\), you have
\[\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty\frac{(n+1)\;n!}{n!}x^n\]

Answer 2

Let's write out our series real quick:$$f(x)=\frac{1!}{0!}+\frac{2!}{1!}x+\frac{3!}{2!}x^2+\cdots$$
Recognize we may condense this series as follows:$$\sum_n\frac{(n+1)!}{n!}x^n=\sum_n(n+1)x^{n}$$
To determine the interval of convergence, we employ the ratio test:$$\begin{align*}C=\lim_{n\to\infty}\left|\frac{(n+2)x^{n+1}}{(n+1)x^{n}}\right|&<1\\\lim_{n\to\infty}\frac{n+2}{n+1}|x|&<1\\|x|&<1&\end{align*}$$
Thus we know our radius of convergence is \(R=1\). We must now test our end-points, \(x=1\) and \(x=-1\) respectively.

Answer 3

@SithsAndGiggles Thanks for your help, but the answer was wrong :/

Answer 4

For \(x=1\), our series reduces to $$\sum_n(n+1)$$Clearly the sequence and therefore diverges (by the n-th term test as \(n+1\to\infty\) as \(n\to\infty\)).

For \(x=-1\), we're left with the alternating series $$\sum_n(-1)^n(n+1)$$
By the alternating series test, however, we see this diverges as (1) \(n+1\to\infty\) as \(n\to\infty\) and (2) \(n+1\not\lt n+2\) so our absolute part is not decreasing.

Answer 5

Thus our our power series fails to converge at the end-points and our interval of convergence is \((-1,1)\).

Answer 6

sequence and therefore *sum* diverge ...

Answer 7

oops, I meant (2) \(n+2\not\leq n+1\).