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As very first task in our analysis I class we got the following task which gives (only) 1 of 16 points. I add the points in order to calculate the needed effort to solve this task...

So let $n,m\in\mathbb{N}$ and $a_n:=\sqrt[n]{b_1^n+...+b_m^n}$, with $b_i\geq 0$ and $i=(1,...,m)$
We have to prove if this sequence converges and determine the limit (by using $\epsilon$ definition).

After running a simple R simulation I found out, that this sequence seems to converge against m.

As I have learned so far, the first step is to determine an $N\in\mathbb{N}$ so that $\forall n\in\mathbb{N}, n\geq N : |a_n-m|<\epsilon$, but while searching for that $N$ I began to struggle.
$|a_n-m|=|\sqrt[n]{b_1^n+...+b_m^n}-m| < \epsilon \Leftrightarrow ... \Leftrightarrow b_1^n+...b_m^n<(\epsilon+m)^n$.
After some research I found, that the LHS can be expressed by polynomials using Faulhaber's formula. At this point I am afraid to be on the wrong track.

Can somebody give me a hint which respects the limitations (having only 1 lecture and it is the first task regarding sequences)?

The R-Code

m_max = 20
m=seq(1,m_max)
sumOfPowers=function(n){sum(m^n)}
n_max = 50
n=seq(1,n_max)
temp=sapply(n,sumOfPowers)
a_n=nthroot(temp,n)
print(data.frame(n,temp,a_n))
Matthias
  • 101

1 Answers1

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Hint

Let $b_k=\max_i b_i$, then$$a_n=b_k\sqrt[n]{\left({b_1\over b_k}\right)^n+\cdots +\left({b_m\over b_k}\right)^n}$$Now prove that $$\sqrt[n]{\left({b_1\over b_k}\right)^n+\cdots +\left({b_m\over b_k}\right)^n}\to 1$$by using $$1\le \sqrt[n]{\left({b_1\over b_k}\right)^n+\cdots +\left({b_m\over b_k}\right)^n}\le\sqrt[n]{m}$$and showing (using the definition of limit) that $$\sqrt[n]{m}\to 1$$

Mostafa Ayaz
  • 33,056