For each n, we define fn:R→R
x↦fn(x)=x2(1+x2)n
We consider the function
f:R→R
x↦f(x)=∞∑n=0fn(x)=∞∑n=0x2(1+x2)n
I want to show that the series does not converge uniformly on [−1,1] but I'm finding it difficult to do that.
First of all, I considered the Weierstrass M test.
MY TRIAL
|fn(x)|=|x2(1+x2)n|≤1(1+x2)n,∀x∈[−1,1],∀n∈N
I'm also thinking that the βn=sup approach could be very helpful too!
Answer
The convergence is not uniform.
For x \ne 0 this is a geometric series:
\sum^{\infty}_{n=0}\frac{x^2}{(1+x^2)^n} = x^2\cdot \frac{1}{1-\frac1{1+x^2}}= x^2+1
and for x = 0 the sum is 0.
Therefore
f(x) = \begin{cases} x^2+1, &\text{ if } x \ne 0\\ 0, &\text{ if }x = 0 \end{cases}
Since f is not continuous, the convergence cannot be uniform.
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