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# Infinite Series

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This might not be the exact place to post this, but there are some excellent mathematicians in here. I beg for a moment of your time. I'm in the process of studying calculus from James Stewart's Calculus: Early Transcendentals Ed 4. One of the problems dealing with infinite series, specifically the ratio test goes as follows... (p739, #35 if anyone has the book handy)
infinity
sigma      (n/n+1)^n^2
n=1

The back of the book says that this is convergent by the ratio test.

However, by the divergence test:
2         2
lim      (n   )n        n
n->inf   (n+1 )     = 1     = 1

1 != 0, thus the series must diverge.

Can someone give me a hand on why the ratio test would make this series converge? When I did the ratio test, I got that the test failed to yield results. I realize that this may smack of someone's homework... but it's really not. The semester is over. I'm looking back over the semester and can't figure this out. Mark (Scout)
All polynomials are funny - some to a higher degree. Furthermore, polynomials of degree zero are constantly funny. [edited by - MrScout on December 14, 2003 2:45:03 AM]

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quote:
Original post by MrScout
                 2         2lim      (n   )n        nn->inf   (n+1 )     = 1     = 1

I don''t see how that limit would reduce to 1.

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Usually, n/(n+1) does go to 1. This is also true when it''s raised to a constant power. But infinity is so big compared to any constant that the power loses all affect. But imagine that the power isn''t constant, but grows with the sequence (by a square power at that) and actually approaches infinity itself. Applying the power separately to both numerator and denominator, the difference between nc and (n+1)c (c being some number "close" to infinity) becomes so astronomically large, that nc becomes nothing compared to (n+1)c, and thus it converges to 0.

The ratio test doesn''t lie, and neither does the book. But I really hate it when books give these types of problems, because things raised to variable powers can be extremely non-trivial, especially for introductory Calculus courses.

I have the same book you are using, but it''s at home

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quote:

James Stewart''s Calculus: Early Transcendentals Ed 4

NOOOOOOOOOES!!!

*hides under calculus book... its big enough anyway...*

hehe anyway i passed it so i dont want to think about it any longer.

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I don''t think the ratio test works in this case either. Does the book work it out?

lim n->inf (n/(n+1) = 1 so it inconclusive by the root test

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polynomials don't affect the ratio test... but it'll go something like this:

                 2              2                   2lim      (n+1)^n    *   (n)  ^n   = lim     (n  )^nn->inf   (n+2)          (n+1)       n->inf  (n+2)

Since limits pass through continuous functions
                2          2(lim     n   )^n   =   1 ^n      = 1 (Ratio test fails)(n->inf  n+2 )

The only solution I can think of is that the solution suggested by Zipster is right.

Which would yield the ratio test:
.....
lim    n^n^2      = 0 (Series is Conv)n->inf (n+1)^n^2

Its not an intro course tho.. its the last Calculus course of my undergraduate math degree. I've dealt with variable powers before... And it just seems to me that ... the limit should goto 1 in the divergence test.

However.. I can see Zipster's point on this one. I just can't see how I could argue it without "hand waving".

Mark (Scout)

All polynomials are funny - some to a higher degree.
Furthermore, polynomials of degree zero are constantly funny.

[edited by - MrScout on December 14, 2003 12:42:33 PM]

[edited by - MrScout on December 14, 2003 12:43:27 PM]

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quote:
Original post by ironpoint
I don''t think the ratio test works in this case either. Does the book work it out?

lim n->inf (n/(n+1) = 1 so it inconclusive by the root test

The root test would make it to the power of n rather than n2, and I still think that would converge to 0.

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The problem is to show that \sum_{n=1}^\infty \left( \frac{n}{n+1} \right)^{n^2} is convergent. First we will look at the terms
a_n = \left( \frac{n}{n+1} \right)^{n^2}. Clearly

\left( \frac{n}{n+1} \right)^{n^2} = \frac{n^n^2}{(n+1)^n^2}

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Oops, let me try that again

Clearly

( n/(n+1) )^n^2 = n^n^2 / (n+1)^n^2
= n^n^2 / n^n^2 + p(n)

where p(n) = (n+1)^n^2 - n^n^2.

It is obvious that p(n) is a non-decreasing function and so
n^n^2 + p(n) dominates n^n^2, thus

lim_{n=1}^\infty ( n/(n+1) )^n^2 =
lim_{n=1}^\infty n^n^2 / n^n^2 + p(n) = 0

srcarrel at uwaterloo dot ca

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This of it this way: as n/(n+1) is always just a little bit less than 1 as n approaches infinity no matter how close it actually gets. From there, it should seem obvious that any number x 0<x<1 raised to infinity will approach 0.

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