Values of x for which series defines a function

[Nemesis10192]

Consider f_m(x)= sum from 1 to infinity of [ (x^n)/(n^m)].

I am asked: for each m in the naturals, find all x in the reals for which the formula f_m defines a function.

But surely for each m in the naturals, any x in the reals defines a function? I don't see why there need be any restrictions. Am I being thick?


The next part asks for m>=2, let x be such that f_m(x) is convergent. Compute f'_m and ensure that you fully justify your answer...


So I was thinking, employ the ratio test to conclude f_m(x) is convergent iff |x|<1. Then check the conditions of some main theorem on differentiability of power series in my notes hold and then I can differentiate in the normal way (since m=/=1).

But mainly the first part...why should there be any restrictions on x? Am I missing something?

 

[HallsofIvy]

This series will "define a function" for a given x only if it give a specific y value for that specific x. In other words, this question is "for what values of x does this series converge?" This is a power series- it will converge in some interval of convergence. And we can determine that by using the "ratio test".

The "ratio test" says that a series \(\displaystyle \sum_{n=0}^\infty a_n\), with \(\displaystyle a_n> 0\) converges if \(\displaystyle \lim_{n \to\infty} \frac{a_{n+1}}{a_n}<1\). Applying the ratio test to \(\displaystyle a_n= \frac{|x|^n}{n^m}\) we have \(\displaystyle \frac{x^{n+1}}{(n+1)^m}\frac{n^m}{|x|^n}= \left(\frac{n+1}{n}\right)^m |x|\). Taking the limit as n goes to 0, \(\displaystyle \lim_{n\to\infty} \frac{n+1}{n}= 1\) so \(\displaystyle \lim_{n\to\infty}\frac{n+1}{n}|x|= |x|\). That is, in order that the power series \(\displaystyle \sum_{n= 0}^\infty \frac{x^n}{n^m}\) converge (and so define a function) we must have |x|< 1.
(Whether the series converges at x= -1 and x= 1, you have to determine separately.)