Re: Evaluate the following limit

In article
<aderamey.addw-8FC39F.21152110082008@xxxxxxxxxxxxxxxxxxxxxx>,
The World Wide Wade <aderamey.addw@xxxxxxxxxxx> wrote:

In article <g7ktd5$6iqi@xxxxxxxxxxxxxxxxxxxx>,
hrubin@xxxxxxxxxxxxxxxxxxxx (Herman Rubin) wrote:

In article <489de860$0$20717$9b4e6d93@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
karl <oudeis@xxxxxxxxxxx> wrote:
T-Hope schrieb:
This has been posted on a forum and is bugging me. I appreciate any
help. I'm sure the OP of the problem would too.



[ infty ]^(1/n)
lim [ ---- n^k ]
n->infinity [ \ ------------------- ]
[ / (k+1)^{k+1} ]
[ ---- ]
[ k=0 ]

-T


Look at it, so nobody can understand it.

Nobody? There is not too much difficulty in doing the
problem, if one knows what is involved in getting an
asymptotic approximation of the sum.

Hint: A sum is approximated by an integral.

At first I didn't quite get what you were hinting at. Is the below
what you had in mind?

Replacing the above series with sum(k=1,oo) n^k/k^k, the resulting
expression is

{ n * sum(k=1,oo) [(k/n)^(-k/n)]^n * (1/n) }^(1/n) (1)

Now n^(1/n) -> 1, so forget about the first n inside the { }. The sum
in (1) looks very much like a Riemann-type sum for an integral. In
other words, it's reasonable to expect (1) to be very close to

{int_[0,oo) [x^(-x)]^n dx}^(1/n) (2)

for n large. By standard L^p theory, (2) approaches the maximum value
of x^(-x) on [0,oo) as n -> oo. Easy calulus shows that max is e^(1/e).

So that's another way to get the limit assuming one can show the
difference between (1) and (2) -> 0 as n -> oo. That shouldn't be too
hard.

To finish up this argument, I'm using ||f||_p -> ||f||_oo provided
some ||f||_p < oo for some p < oo; this is true on any measure space.
Below we have the requirement ||f||_p < oo in spades.

Set f(x) = x^(-x), and let f_n be the step function sum(k=1,oo)
f(k/n)*X_[(k-1)/n, k/n). Then f_n -> f uniformly on [0,oo), and
because f is decreasing on [1/e, oo), f_n < f on [1,oo). Note
||f_n||_n is exactly the sum in (1) above.

Let eps > 0 be small. Define g(x) = eps on [0,1], g(x) = 0 for x > 1.
Then for large n, f_n < f + g. So

lim sup ||f_n||_n <= lim ||f + g||_n

= ||f + g||_oo = e^(1/e) + eps.

Also,

lim inf ||f_n||_n >= lim inf (int_[0,1] f_n^n)^(1/n)

>= lim [int_[0,1] (f-g)^n]^(1/n) = e^(1/e) - eps.

Because eps is arbitrary, ||f_n||_n -> e^(1/e) as desired.




lim ||f + g||_n
.

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