# Thread: Gauss Test for Convergence

1. ## Gauss Test for Convergence

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.

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 $\approx$).

2. ## Re: Gauss Test for Convergence

Originally Posted by process91
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.

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 $\approx$).

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

... we obtain...

$\ln \frac{a_{n+1}}{a_{n}}= k\ \{\ln (1+\frac{1}{2n})-\ln (1+\frac{1}{n})\} = k\ \{-\frac{1}{2n} + \text{o}(\frac{1}{n})\}$ (2)

... so that is...

$\frac{a_{n+1}}{a_{n}}= e^{-k \{\frac{1}{2n}+ \text{o} (\frac{1}{n})\}}= 1 - \frac{k}{2n} + \text{o} (\frac{1}{n})$ (3)

Now applying the Gauss' test we find that the series...

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

... converges for $k>2$ and diverges for $k\le 2$...

Kind regards

$\chi$ $\sigma$

3. ## Re: Gauss Test for Convergence

Thank you very much for your reply. I'm still a little confused; either the book has not yet covered anything to explain the steps in both (2) and (3), or I missed it (this is Calculus Vol I. by Apostol). It did cover little-o notation in the context of Taylor Series, but that was only useful at the time because we were considering $n \rightarrow a$. I see that

$\lim_{n \rightarrow \infty} \frac{\ln(1+\frac{1}{2n}) - \ln(1+\frac{1}{n})+\frac{1}{2n}}{\frac{1}{n}}=0$, so I understand how the statement is true, but have no idea how to have come up with attempting that on my own. Is this a known result that one will apply? Where could I read more about this?

In step (3), I am completely at a loss for how to proceed. I don't know what to do with the little-o in the exponent, and I'm not sure how you got your result.

I think that the answer to this question requires external knowledge about little-o notation that I do not have yet. I don't want to take up so much of your time, if you would be so kind as to point me to another reference I would appreciate it.

,

,

,

### gauss test of convergence

Click on a term to search for related topics.