How do you prove that the series {n^(1/n)} from n=3 to infinity is decreasing?

Question

Is there a proof by induction?

Answer 1

This is interesting, finding a short way to prove this, but here it goes:

We know that e = (1+x)^1/x in the limiting case x->0. From this, we know that

e^x > 1+x

for all positive real x. From this, we can say that

(e+y)^x > e^x > 1+x

for all positive real x, y. Now, making a subsitution x = k/(e+y), we have

(e+y)^(k/(e+y)) > 1 + (k/(e+y))

for all positive real k, y. Multiplying both sides by (e+y), we have

(e+y)^(k/(e+y)+1) > e+y+k

Raising both sides by power 1/(e+y+k), we have

(e+y)^(1/(e+y)) > (e+y+k)^(1/(e+y+k)

for all positive real y, k. Thus, for any value n = e+y, we know that

n^(1/n) > (n+k)^(1/(n+k))

and thus it's proven, without resorting to calculus.

Answer 2

Let y = n^(1/n)
lny = ln(n) / n
Differentiate
1/y * dy/dn = (1 - ln(n)) / n^2
dy/dn = o when ln(n) = 1
or when n = e= 2.718

Differentiate again:
-1/y^2 * dy/dn + 1/y * d2y/dn2 = (-n - 2n + 2nln(n))/n^4
when n = e
0 + e^(1/e) y" = -1/e^3
ie y" = -ve
So y is MAX when n = e

So for n >= 3 the function is decreasing.

Answer 3

this question just made me realize i forgot everything i learnt in calculus last term.. oh well...

looking at that term, i think you've to take a log somewhere. Is there a log test? i remember it vaguely.