What is the limit of the infinite sum:
[symbol]S[/symbol](n[sup]1/n[/sup] - 1)?
Is there an “easy”* way to prove it?
(*with “easy” defined as needing only undergraduate math)
Thanks.
Well, Mathematica tells me it diverges.
There are lots of convergence tests . . . however the ones I’m remembering don’t seem very useful in this case.
I can tell you offhand it diverges. Still working on how to prove it with undergrad math, preferably Calc 2 techniques. The best idea I have so far is an integral test comparing with \integral_1^\infinity(x^{1\over x}-1)\,dx, then showing that that’s a slower convergence than something like 1/x, but exactly how escapes me at the moment. Damned analysis.
The root test might help:
http://oregonstate.edu/dept/math/CalculusQuestStudyGuides/SandS/SeriesTests/root.html
Even to me, as (very) non-mathematician, it looks divergent. Doesn’t it break down into [symbol]S[/symbol]n[sup]1/n[/sup] - [symbol]S[/symbol]1? And, doesn’t the second term simply equal negative infinity? So the only way it could converge would be for the whole expression to equal zero, and the the first term would have to equal infinity.
But again, I’m a very-nonmathematician, so take this with a large rock of salt.
Just because two sequences diverge does not mean that their difference diverges. For example,
1, 2, 3, 4, 5, …, n
and
1.1, 2.01, 3.001, 4.0001, 5.00001, … , n+(1/10^n)
clearly both diverge, but their difference
0.1, 0.01, 0.001, 0.0001, 0.00001, …, 1/10^n
does not.
It seems like it should be possible to show that the terms are bounded below by 1/n, but I’m not having much luck.
It wasn’t convinced that it diverged since each term (n[sup]1/n[/sup] - 1) seems to be getting increasingly smaller as n increases.
eg. the 1000th root of 1000 minus 1 is 1.00693 - 1 = 0.00693…
And the 1000000th root of 1000000 minus 1 is 1.0000138 - 1 = 0.0000138…
Mostly I was wondering how to prove if it converges or diverges (and if the former, to what).
Thanks to all who have contributed.
That should read, “I wasn’t convinced …”.
Just thought I’d chime in that not every decreasing series will converge. The most famous example of this is the harmonic series:
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12
Obviously decreases with each term, and just as obviously you can find a term as close to 0 as you want to.
But the sum of the series has no limit. Can’t remember exactly how you prove that. 
Doh… that should be “+ 1/11 + 1/12 + …”
Repeat, preview is my friend…
That’s a good point. If you can show that n[sup]1/n[/sup] - 1 decreases faster than 1/n, then Σ(n[sup]1/n[/sup] - 1) doesn’t converge.
If this is true, that means for some constants C and k, that C/n > n[sup]1/n[/sup] - 1 for all n > k. For that to be true, then the derivative of C/n would be greater than the derivative of n[sup]1/n[/sup] - 1 for all n > k.
The derivative of C/n is -C/n[sup]2[/sup], and the derivative of n[sup]1/n[/sup] - 1 is n[sup]1/n[/sup] * (-1/n[sup]2[/sup] * ln(n) + 1/n[sup]2[/sup]), so -C would be greater than n[sup]1/n[/sup](1 - ln(n)) for all n > k. Now n[sup]1/n[/sup] will always be a positive number, and (1 - ln(n)) will always be a negative number when n > e, so for any number C where C < 0, -C > n[sup]1/n[/sup](1 - ln(n)) for n > e. Therefore, Σ(n[sup]1/n[/sup] - 1) doesn’t converge.
What do you think, sirs?
Whoops. The derivative of n[sup]1/n[/sup] - 1 is (n[sup]1/n[/sup] - 1) * (-1/n[sup]2[/sup] * ln(n) + 1/n[sup]2[/sup]). But n[sup]1/n[/sup] - 1 will always be a positive number for n > 1.
The series is divergent. Consider the tail of the series, starting at n=3>e. Now clearly
Σ(n[sup]1/n[/sup]-1) > Σ(e[sup]1/n[/sup]-1) > Σ(1/n),
divergent. (Apologies if my Sigmas don’t show up right; I don’t have a normal Symbol font here.)
By comparison. Consider the series S[sub]2[/sub] = 1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + … Clearly, this series is less than the harmonic series, since each term is less than or equal to the corresponding term in the harmonic series. But S[sub]2[/sub] = (1) + (1/2 + 1/2) + (1/4 + 1/4 + 1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8) + …
= 1 + 1 + 1 + 1 + …, which is trivially divergent.
Very elegant! Thank you.
I’m sure I’m missing something obvious here–how does this work?
e[sup]x[/sup] = 1 + x + x[sup]2[/sup]/2! + x[sup]3[/sup]/3! …
e[sup]1/n[/sup] - 1 = 1/n + 1/2n[sup]2[/sup] + … > 1/n
Alternately, for x>0 we have (since e[sup]x[/sup] is strictly increasing)
x = x e[sup]0[/sup] < ∫[sub]0[/sub][sup]x[/sup] e[sup]t[/sup] dt = e[sup]x[/sup]-1 .
(That symbol after the < is supposed to be an integral, in case it doesn’t show up right.)