Question

Can someone help me perform the ratio test of convergence to the sum of (k^20)(x^k)/(2k+1)!

Answer 1

Here, you want to take
\[\large a_k = \frac{k^{20}x^k}{(2k+1)!}\]
Then we see that
\[\large a_{k+1} = \frac{(k+1)^{20}x^{k+1}}{(2(k+1)+1)!} = \frac{(k+1)^{20}x^{k+1}}{(2k+3)!}\]
Now, the series converges by ratio test if
\[\large L = \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| < 1\]
Using our values for \(\large a_k\) and \(\large a_{k+1}\), we have that
\[\large \begin{aligned} \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty}\left|\frac{(k+1)^{20}x^{k+1}}{(2k+3)!}\cdot\frac{(2k+1)!}{k^{20}x^k}\right|\\ &= \lim_{k\to\infty}\left|\frac{(k+1)^{20}\color{purple}{(2k+1)!}\color{green}{x^k}\cdot x}{(2k+3)(2k+2)\color{purple}{(2k+1)!}k^{20}\color{green}{x^k}}\right| \\ &= \lim_{k\to\infty} \left|\frac{(k+1)^{20}}{(2k+3)(2k+2)k^{20}}x\right|\\ &= \left|\lim_{k\to\infty}\left(\frac{(k+1)^{20}}{(2k+3)(2k+2)k^{20}}\right) x\right|\end{aligned}\]
Now, to evaluate the limit, compare the degrees of the numerator and denominator of the k terms (here, x is nothing more than a constant since the limit is taken with respect to k).

Once you evaluate this limit, what conclusion can be drawn about convergence?

I hope this makes sense! :-)

Answer 2

so to clarify (2k+3)!=(2k+3)(2k+1)k! ? And your limit is zero correct?

Answer 3

(2k+3)! = (2k+3)(2k+2)(2k+1)!

And yes, the limit is zero. So the series converges.

Bonus question: What's the radius and interval of convergence?

Answer 4

radius is infinity and interval of convergene is (-inf,inf). But what does (2k+1)! simplify to??

Answer 5

What do you mean by simplifying (2k+1)! ?

Using the fact that n! = n(n-1)(n-2)(n-3) ... (3)(2)(1), then it would follow in a similar fashion that (2k+1)! = (2k+1)(2k)(2k-1)(2k-2)...(k+2)(k+1)k(k-1)(k-2)...(3)(2)(1).

Does this clarify things?

Answer 6

And yes, whenever you get a limit of zero in the ratio test, that generally implies that you have infinite radius and interval of convergence.