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#### Chipset3600

##### Member

- Feb 14, 2012

- 79

Curious to me is that the the ratio and root test have the same conditions.

How can i basically prove this equality?

\(\displaystyle \frac{a_{n+1}}{a_{n}} = \sqrt[n]{a_{n}}\)

Thank you!

- Thread starter Chipset3600
- Start date

- Thread starter
- #1

- Feb 14, 2012

- 79

Curious to me is that the the ratio and root test have the same conditions.

How can i basically prove this equality?

\(\displaystyle \frac{a_{n+1}}{a_{n}} = \sqrt[n]{a_{n}}\)

Thank you!

- Aug 18, 2013

- 76

So what is the sequence ${a_n}$?

- Thread starter
- #3

- Feb 14, 2012

- 79

There is no sequence, is just to prove...So what is the sequence ${a_n}$?

- Thread starter
- #4

- Feb 14, 2012

- 79

is a rule of sequences that I found.

If i use this rule in this exercise:

And apply limit in booth sides i will have the same result. But i want to know why this property is true...

If i use this rule in this exercise:

And apply limit in booth sides i will have the same result. But i want to know why this property is true...

Last edited:

- Sep 16, 2013

- 337

Curious to me is that the the ratio and root test have the same conditions.

How can i basically prove this equality?

\(\displaystyle \frac{a_{n+1}}{a_{n}} = \sqrt[n]{a_{n}}\)

Thank you!

Hi Chipset!

My knowledge is a bit sketchy as far as series summation techniques, and if I've got this wrong, then hopefully some kind soul on here will take pity on both of us and correct me, but here goes...

Firstly, with regards to both the Root Test and Ratio Test, you need to bear in mind that we are talking about the limiting value of a series term, or the limiting value of a quotient of consecutive series terms. So, for example, if we have a series of the form:

\(\displaystyle \sum_{k=0}^{\infty}a_k\)

then if the limit

\(\displaystyle \lim_{k\to\infty}\, \Bigg|\frac{a_{k+1}}{a_k}\Bigg|=r\)

exists, the series is absolutely convergent for \(\displaystyle r < 1\), divergent for \(\displaystyle r > 1\), and of undetermined convergence/divergence if \(\displaystyle r = 1\)

Similarly, for the Root Test, if the limit

\(\displaystyle \lim_{k\to\infty}\, \Bigg|\sqrt[k]{a_k}\Bigg|=r\)

exists, then once again, the series in question is absolutely convergent for \(\displaystyle r < 1\), divergent for \(\displaystyle r > 1\), and of undetermined convergence/divergence if \(\displaystyle r = 0\).

This is the bit where I'm not sure, but I suspect that you're approaching it the wrong way by assuming equality for

\(\displaystyle \frac{a_{n+1}}{a_{n}} = \sqrt[n]{a_{n}}\)

in the general sense. In the limit, as \(\displaystyle n\to\infty\), then equality should hold, but for a general, arbitrary series term, this is unlikely to be true.

Can anyone add to or correct that...??? Please and thankuppo!

EDIT:

Apologies... I missed that last post of yours...

For a

if \(\displaystyle \lim_{n\rightarrow\infty}{a_{n+1}\over a_n}\) exists, then \(\displaystyle \lim_{n\rightarrow\infty}a_n^{1/n}\) exists and the two limits are equal.

However, it's possible that \(\displaystyle \lim_{n\rightarrow\infty}a_n^{1/n}\) exists, but \(\displaystyle \lim_{n\rightarrow\infty}{a_{n+1}\over a_n}\) does not exist!

Example: let the sequence {a

The ratio and root tests have "generalizations", but this is more in the realm of advanced calculus. These tests involve the limit supremum and limit infimum of sequences. If you're up to it,

you can find a wealth of information about this on the web.