Prove Convergence of Series Using Gauss' Test

[process91]

Homework Statement

I just got done proving Gauss' test, which is given in the book as:

If there is an $N\ge 1$, an $s>1$, and an $M>0$ such that
$$\frac{a_{n+1}}{a_n}=1 - \frac{A}{n} + \frac{f(n)}{n^s}$$
where $|f(n)|\le M$ for all n, then $\sum a_n$ converges if $A>1$ and diverges if $A \le 1$.

This is equivalent to the many other forms I have found on the web. The next question asks to use this test to prove that the series

$$\sum_{n=1} ^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)} \right)^k$$

converges if $k>2$ and diverges if $k \le 2$ using Gauss' test.

Homework Equations

The Attempt at a Solution

OK, so much for the preamble. Here's my attempt:

$$\frac{a_{n+1}}{a_n} = \left ( \frac {2n+1}{2n+2} \right ) ^ k$$

And that's it... I don't know how to put this into a form which corresponds in general to the form required for Gauss' test. I saw a similar problem online solved with the use of the fact that $\left (\frac{n}{n+1} \right)^k \approx 1 - \frac{k}{n}$ for large n, but the book has not covered anything like that (it introduced the ~ symbol, but not $\approx$).

 

[mighty2000]

Thank you for sharing your thoughts on using Gauss' test to prove the convergence of the given series. It seems like you are on the right track, but there are a few adjustments that can be made to better fit the form required for Gauss' test.

Firstly, we can rewrite the given series as:

\sum_{n=1} ^{\infty} \left( \frac{(2n-1)!!}{(2n)!!} \right)^k

where (2n-1)!! represents the double factorial of (2n-1) and (2n)!! represents the double factorial of (2n). This allows us to simplify the expression to:

\frac{a_{n+1}}{a_n} = \left ( \frac{2n+1}{2n+2} \right ) ^ k \approx \left ( \frac{n}{n+1} \right ) ^ k

as n becomes large. This approximation is valid because the terms inside the parentheses approach 1 as n becomes large, which fits the requirement of |f(n)|\le M for all n.

Now, we can rewrite the approximation as:

\left ( \frac{n}{n+1} \right ) ^ k = \left ( 1 - \frac{1}{n+1} \right ) ^ k \approx 1 - \frac{k}{n+1}

Again, this approximation holds as n becomes large and fits the form required for Gauss' test.

Finally, we can rewrite the expression as:

\frac{a_{n+1}}{a_n} = 1 - \frac{k}{n+1}

which now matches the form required for Gauss' test with A = k and f(n) = 0. Therefore, using Gauss' test, we can conclude that the given series converges if k > 1 and diverges if k \le 1.

I hope this helps and clarifies any confusion you may have had. Keep up the good work in exploring mathematical concepts and their applications.