# Clarification of Root Test & Ratio Test

• Jul 27th 2010, 07:39 PM
TaylorM0192
Clarification of Root Test & Ratio Test
I read in a text book recently that when testing an infinite series for convergence, if the ratio test fails (r = 1), to not try and attempt using the root test, as it too will necessarily fail.

However, I think this is wrong, and I just want to confirm it here.

I was working this series: $\displaystyle \Sigma \frac{n^3}{5^n}$ for which the ratio test fails and the root test gives $\displaystyle \frac{1}{5}$ implying convergence for this positive-termed series.

Maybe I did it incorrectly - so you should double check if you don't agree with me.
• Jul 27th 2010, 08:06 PM
bondesan
$\displaystyle \displaystyle{\lim_{n\to\infty}\left|\frac{a_{n+1} }{a_n}\right|}$

$\displaystyle \displaystyle{\lim_{n\to\infty}\left|\frac{(n+1)^3 }{5^{n+1}}\cdot\frac{5^n}{n^3}\right|}$

$\displaystyle \displaystyle{\lim_{n\to\infty}\left|\frac{n^3 + 3n^2 + 3n + 1}{5n^3}\right| = 1/5}$

The ratio test works too.
• Jul 27th 2010, 08:27 PM
Failure
Quote:

Originally Posted by TaylorM0192
I read in a text book recently that when testing an infinite series for convergence, if the ratio test fails (r = 1), to not try and attempt using the root test, as it too will necessarily fail.

That's not true, the root test is somewhat stronger than the ratio test. An example for which the ratio test fails, whereas the root test succeeds in showing convergence, would be

$\displaystyle \sum_{k=0}^\infty\left(\frac{1}{2}\right)^{k+(-1)^k}$

Quote:

However, I think this is wrong,
Right.

Quote:

and I just want to confirm it here.
I was working this series: $\displaystyle \Sigma \frac{n^3}{5^n}$ for which the ratio test fails and the root test gives $\displaystyle \frac{1}{5}$ implying convergence for this positive-termed series.
No, that example does not work, because even in this case the ratio test succeeds

$\displaystyle \frac{(n+1)^3/5^{n+1}}{n^3/5^n}=\frac{\left(1+\frac{3}{n}+\frac{3}{n^2}+\frac {1}{n^3}\right)}{5}\rightarrow \frac{1}{5}$
• Jul 28th 2010, 07:19 AM
roninpro
I think that the statement in the original post needs to be qualified a little bit. We have the following inequality:

$\displaystyle \liminf \frac{a_{n+1}}{a_n}\leq \liminf \sqrt[n]{a_n}\leq \limsup \sqrt[n]{a_n}\leq \liminf \frac{a_{n+1}}{a_n}$

We are told that $\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=1$. But then, $\displaystyle \liminf \frac{a_{n+1}}{a_n}=\limsup \frac{a_{n+1}}{a_n}=1$ which forces $\displaystyle \lim_{n\to \infty} \sqrt[n]{a_n}=1$ as well.

So, I think that the book's statement is correct. If the limit of the ratio exists, then the limit of the roots will exist and converge to the same number. In particular, if the ratio test gives limit 1, then the root test will too.
• Jul 28th 2010, 07:45 AM
Ted
$\displaystyle \dfrac{n^3}{5^n} \leq \left(\dfrac{4}{5}\right)^n$