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Sum series- Prove the equality of ratio and root.

Chipset3600

Member
Feb 14, 2012
79
I found this on the internet, but did not find the proof.
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!
 

eddybob123

Active member
Aug 18, 2013
76
So what is the sequence ${a_n}$?
 

Chipset3600

Member
Feb 14, 2012
79

Chipset3600

Member
Feb 14, 2012
79
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
I found this on the internet, but did not find the proof.
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! :D


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! (Inlove)


EDIT:

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

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Hi,
For an a sequence of positive terms, the following is true:

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 {an} be 1, 1/2, 1, 1/2 ... Then clearly the ratio of successive terms alternates between 2 and 1/2, so the limit of this ratio can't exist. But the nth root of any positive is very close to 1 provided n is sufficiently large. So the nth root of an does approach 1 as its limit.

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.