Ratio test for series divergence help.

**chsoviz0716** · December 14th 2011, 10:11 PM

This is from so called ratio test, which says

The series Ʃa(n) diverges if abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer.

I was wondering if I could replace the condition with lim inf abs{a(n+1)/a(n)} ≥ 1.

So basically what I'm asking is whether these two below are equivalent or not.
1. abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer
2. lim inf abs{a(n+1)/a(n)} ≥ 1

*a(n) means nth term of the sequence, and abs means absolute value.

Thank you in advance.

**chisigma** · December 15th 2011, 01:10 AM

I think You couldn't. Supposing that is...

$\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}}|=1$ (1)

... the behaviour of $|\frac{a_{n+1}}{a_{n}}}|$ can be 'oscillating' around 1 and in this case it isn't $|\frac{a_{n+1}}{a_{n}}}| \ge 1$ 'for $n \ge N$ where N is some fixed integer'...

Marry Christmas from Serbia

$\chi$ $\sigma$

**Prove It** · December 15th 2011, 02:16 AM

Originally Posted by chsoviz0716

This is from so called ratio test, which says

The series Ʃa(n) diverges if abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer.

I was wondering if I could replace the condition with lim inf abs{a(n+1)/a(n)} ≥ 1.

So basically what I'm asking is whether these two below are equivalent or not.
1. abs{a(n+1)/a(n)} ≥ 1 for n≥N where N is some fixed integer
2. lim inf abs{a(n+1)/a(n)} ≥ 1

*a(n) means nth term of the sequence, and abs means absolute value.

Thank you in advance.

Actually, the series is only convergent where $\displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \end{align*}$ and divergent where $\displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| > 1\end{align*}$ .

The ratio test is INCONCLUSIVE if $\displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| = 1 \end{align*}$ .

**chisigma** · December 15th 2011, 02:36 AM

The series diverges if $\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}|=1$ but is also $\forall{n>N}\ |\frac{a_{n+1}}{a_{n}}|\ge 1$ ...

Marry Christmas from Serbia

$\chi$ $\sigma$

**chsoviz0716** · December 15th 2011, 02:54 AM

Originally Posted by chisigma

I think You couldn't. Supposing that is...

$\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}}|=1$ (1)

... the behaviour of $|\frac{a_{n+1}}{a_{n}}}|$ can be 'oscillating' around 1 and in this case it isn't $|\frac{a_{n+1}}{a_{n}}}| \ge 1$ 'for $n \ge N$ where N is some fixed integer'...

Marry Christmas from Serbia

$\chi$ $\sigma$

Thank you for the reply first.
The example you took shows even if 2 is satisfied, 1 need not be satisfied, so clearly they are not equivalent.
What about the other way around?
Can I conclude that if 1 is satisfied, 2 is satisfied?
or is there an example which satisfies 1 but doesn't satisfy 2?

**chisigma** · December 15th 2011, 03:15 AM

Can I conclude that if 1 is satisfied, 2 is satisfied?... or is there an example which satisfies 1 but doesn't satisfy 2? ...

Let's indicate $\rho_{n}= \frac{a_{n+1}}{a_{n}}$ and consider the possibility $\rho_{n}=2+(-1)^{n}$ . Clearly 1) is satisfied and 2) is not because the limit for n tending to infinity of $\rho_{n}$ doesn't exist...

Marry Christmas from Serbia

$\chi$ $\sigma$

**chsoviz0716** · December 15th 2011, 03:51 AM

Originally Posted by chisigma

Can I conclude that if 1 is satisfied, 2 is satisfied?... or is there an example which satisfies 1 but doesn't satisfy 2? ...

Let's indicate $\rho_{n}= \frac{a_{n+1}}{a_{n}}$ and consider the possibility $\rho_{n}=2+(-1)^{n}$ . Clearly 1) is satisfied and 2) is not because the limit for n tending to infinity of $\rho_{n}$ doesn't exist...

Oh may be you misunderstood what I wrote as 'lim inf'. I didn't mean limit as n goes to infinity but meant 'limit inferior' which is defined to be the infimum of all subsequential limits of the original sequence. I attach the link here.
Limit superior and limit inferior - Wikipedia, the free encyclopedia

So even in the case of 2+(-1)^n, lim inf abs{a(n+1)/a(n)} ≥ 1 still holds. In fact, it equals one in this case.

**chisigma** · December 15th 2011, 06:16 AM

For the sequence $\rho_{n}=\frac{|a_{n+1}|}{|a_{n}|}= 2+(-1)^{n}$ it doesn't exist $\lim_{n \rightarrow \infty} \rho_{n}$ ...

Marry Christmas from Serbia

$\chi$ $\chi$

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