What if a ratio test of series = 1

May 2009
71
0
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?
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
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
 
Jul 2009
555
298
Zürich
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:

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

The first factor goes to 1 as n goes to \(\displaystyle \infty\), since

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

and the second factor goes to 0.


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.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
No, it doesn't. The ratio test acually tells you that this series converges absolutely:

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

The first factor goes to 1 as n goes to \(\displaystyle \infty\), since

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

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 \(\displaystyle \frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}\) go to \(\displaystyle 1 \) ?

We can use this inequality

\(\displaystyle \ln{n} < n < n+1 \)


If a ratio test of series = 1 , it means nothing .
 
Jul 2009
555
298
Zürich
Doesn't it \(\displaystyle \frac{\ln(n+1)}{\ln (n)}\cdot\frac{n^3}{n^3+3n^2+3n+1}\) go to \(\displaystyle 1 \) ?
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 \(\displaystyle \sum_n\frac{1}{n^2}\) converges and can be used to bound the given series from above.)

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.
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
Try using the fact that for \(\displaystyle n > 0 \) that

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

(and read the previous post).
 
May 2009
71
0
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.