uniform convergence of $S_n=n^2x/(1+n^3x^2)$

Question

Comment about the uniform convergence of the sequence, $S_n(x)=n^2x/(1+n^3x^2)$ , over $[0,1].$

I found that the series is pointwise convergent to $0$ for all $x$. But for showing uniform convergence or not, I tried to use the definition $∀\epsilon>0$: $∃ m∈N$ such that $|f_n(x)−f(x)|<ϵ \ \ \forall n>m(ϵ)$ .

I need help to solve this using epsilon-delta definition as I can easily solve it using test for uniform convergence.

Any hint would be highly appreciated.

Answer 1

If you mean the Weierstrass M-test , you should be careful about the differences between the theorem's statement and your own problem as in your case we are not dealing with sums.

However, please note that for every $n \in \mathbb N$, $0<\frac{1}{n^{\frac{3}{2}}} \leq 1$ and:

$$|S_n(\frac{1}{n^{\frac{3}{2}}})-0|=\frac{n^2 \times \frac{1}{n^{\frac{3}{2}}}}{1+n^3 \times (\frac{1}{n^{\frac{3}{2}}})^2}=\frac{\sqrt n}{2} \geq \frac{1}{2}.$$

So, the sequence is not uniformly convergent to $0$.

If you wonder how we can usually find a counterexample showing the sequence is not uniformly convergent, the derivative could be a possible approach. For example in your case, we have:

$$S'(x)=0 \implies x^2=\frac{1}{n^3}.$$ More examples are available here, using the same approach.