[Guest]

If a>b>c...>0 for an arbitrary number of numbers, what is the limit as n approaches infinity of the nth root of the sum a^n+b^n+c^n+... using all of the same numbers?

I searched for this and did not see it posted, but I could have missed it.

This really has a simple answer, but the proof is somewhat complex.

[Guest]

Having done a quick calculation, it seems that the answer is always 'a'. For n=0, you obviously have a+b+c.... which tends pretty quickly back to 'a' as n-> infinity.

I would guess the proof requires some expansion of:

aⁿ + (a-i)ⁿ + (a-j)ⁿ ...... [where i, j, etc > 0 ]

= r.aⁿ + n.a^n-1.(i+j+...) - ¹/₂n.(n+1).a^n-2.(i+j+...)²... [where r is the number of terms used originally]

and shows that the remaining terms tend to (r-1).aⁿ (probably iteratively)

[Guest]

If a>b>c...>0 for an arbitrary number of numbers, what is the limit as n approaches infinity of the nth root of the sum a^n+b^n+c^n+... using all of the same numbers?

I searched for this and did not see it posted, but I could have missed it.

This really has a simple answer, but the proof is somewhat complex.

a, the biggest number?

As n goes to infinity, aⁿ gets larger more quickly than any number smaller than it.

[bonanova]

If a>b>c...>0 for an arbitrary number of numbers, what is the limit as n approaches infinity of the nth root of the sum a^n+b^n+c^n+... using all of the same numbers?

a.

To see this, write aⁿ+bⁿ+cⁿ+... as aⁿ x [1+(b/a)ⁿ+(c/a)ⁿ+...].

The terms b/a, c/a, ... are less than unity.

Raising them to arbitrarily high power creates vanishingly small values.

Thus as n approaches infinity the value of the square bracket approaches unity.

That leaves you with the n^th root of aⁿ which is simply a.

[Guest]

a.

To see this, write aⁿ+bⁿ+cⁿ+... as aⁿ x [1+(b/a)ⁿ+(c/a)ⁿ+...].

The terms b/a, c/a, ... are less than unity.

Raising them to arbitrarily high power creates vanishingly small values.

Thus as n approaches infinity the value of the square bracket approaches unity.

That leaves you with the n^th root of aⁿ which is simply a.

You always have the easy answer, don't you!? Nicely done!

Note to self: when bonanova is 'posting....' , wait before trying to post your own clever answer (which doesn't actually work)

[Guest]

Most people got that the answer was "a." This is called the "infinity-norm." The general term with n is the n-norm. If a specific number is put in, as in 2, it is the 2-norm, also known as the Euclidean norm. So as n approaches infinity, it takes the name infinity-norm. I thought the proof from Bonanova was very nice and compact. Thanks.