# What if a ratio test of series = 1

• Jun 1st 2010, 02:42 AM
orendacl
What if a ratio test of series = 1
SUM, from 1 to infinity of:

ln(n)
________
(n+1)^3

I work this out using the ratio test for convergence and the answer I get is "1"...

Does this sound right?
If so, this then means that there are no other tests that can be done correct? The test is simply inconclusive.
Or:
Is there another test that should be performed next?
• Jun 1st 2010, 03:16 AM
Prove It
Quote:

Originally Posted by orendacl
SUM, from 1 to infinity of:

ln(n)
________
(n+1)^3

I work this out using the ratio test for convergence and the answer I get is "1"...

Does this sound right?
If so, this then means that there are no other tests that can be done correct? The test is simply inconclusive.
Or:
Is there another test that should be performed next?

You could try the root test...

Root test - Wikipedia, the free encyclopedia
• Jun 1st 2010, 03:20 AM
Failure
Quote:

Originally Posted by orendacl
SUM, from 1 to infinity of:

ln(n)
________
(n+1)^3

I work this out using the ratio test for convergence and the answer I get is "1"...

Does this sound right?

No, it doesn't. The ratio test acually tells you that this series converges absolutely:

$\frac{a_{n+1}}{a_n}=\frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}\rightarrow 0$

The first factor goes to 1 as n goes to $\infty$, since

$\frac{\ln(n+1)}{\ln (n)}=\frac{\ln(n)+\ln(1+1/n)}{\ln(n)}$

and the second factor goes to 0.

Quote:

If so, this then means that there are no other tests that can be done correct?
No: if the ratio test fails, the root test might still be used to prove absolute convergence.
• Jun 1st 2010, 04:33 AM
simplependulum
Quote:

Originally Posted by Failure
No, it doesn't. The ratio test acually tells you that this series converges absolutely:

$\frac{a_{n+1}}{a_n}=\frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}\rightarrow 0$

The first factor goes to 1 as n goes to $\infty$, since

$\frac{\ln(n+1)}{\ln (n)}=\frac{\ln(n)+\ln(1+1/n)}{\ln(n)}$

and the second factor goes to 0.

No: if the ratio test fails, the root test might still be used to prove absolute convergence.

Doesn't it $\frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}$ go to $1$ ?

We can use this inequality

$\ln{n} < n < n+1$

If a ratio test of series = 1 , it means nothing .
• Jun 1st 2010, 05:15 AM
Failure
Quote:

Originally Posted by simplependulum
Doesn't it $\frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}$ go to $1$ ?

Oops, yes, you are right (obviously I didn't even take the time to read carefully enough what I had written myself).

Maybe it was because I was already quite sure that the series converges. (Why? - Because I know that $\sum_n\frac{1}{n^2}$ converges and can be used to bound the given series from above.)

Quote:

If a ratio test of series = 1 , it means nothing .
Sure, in that case, the ratio test does not decide the question of convergence vs. divergence.
• Jun 1st 2010, 05:58 AM
Jester
Try using the fact that for $n > 0$ that

$
\frac{\ln n}{(n+1)^3} < \frac{\ln n}{n^3} < \frac{n}{n^3} = \frac{1}{n^2}
$

(and read the previous post).
• Jun 1st 2010, 10:16 AM
orendacl
OK,
I said that initially the test was inconclusive as via Ratio = 1, then using the SUM fron n to infinity of 1/n^2 as being a bound for the problem which "is" convergent, thus this is a CONVERGENT series.