Radius of convergence?

Question

Hi,

I am trying to find the radius of convergence for [(2n)!x^n] / (n!)^2. I did the ratio test and have it down to:

[(2n+2)! * x * (n!)^2] / [(2n)! * ((n+1)!)^2].

I was wondering if I could simplify [(2n+2)!] / (2n)! to 1? If so, could I then simplify [(n)!]^2 / [(n+1)!]^2 to 1 as well?


Thanks!

Answer 1

You are correct in attempting to use the Ratio Test. Let me walk you through the steps I took so you can compare your answer to mine.

The first step simply is the set up of the Ratio Test:

Lim as n-->infinity of [ (2[n+1])! * x^(n+1) * (n!)^2 ] / [ ([n+1]!)^2 * (2n)! * x^n ]

By splitting up the more complex exponents and some simple algebra, like distributing terms, we can now write out the expression as:

Lim as n-->infinity of [ (2n + 2)! * x^n * x * (n!)^2 ] / [ ([n+1]!)^2 * x^n * (2n)! ]

Notice that in the numerator and denominator that we can cancel out x^n, giving us:

Lim as n-->infinity of [ (2n + 2)! * (n!)^2 * (x) ] / [ ([n+1]!)^2 * (2n)! ]

Notice that now in the numerator we have (2n + 2)! and in the denominator we have (2n)!. As n goes to infinity, these quantities become nearly the same. Thus, we can simplify the expression by canceling them out.

Also notice that now in the numerator we have (n!)^2 and in the denominator we have ([n+1]!)^2. As n goes to infinity, these quantities also become nearly the same. Thus, we can simplify the expression again by cancelling these two terms out.

Now, our expression is left to be:

Lim n-->infinity of [ x ]

which we of course know is [ x ].

For the series to converge, the absolute value of x has to be less than 1; however, we are not done yet because we need to test the endpoints (which are 1 and -1).

If we plug 1 into the original equation for x, our expression becomes 2n. This series diverges because you would just be infinitely adding terms together. Thus, the endpoint of 1 is not included in the interval of convergence.

If we plug -1 into the original equation for x, our expression is non-existent since we cannot take the factorial of a negative number. Thus, -1 does not work and is not included in the interval of convergence.

Therefore, our final answer is:

The series converges for -1 < x < 1