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=supx∈[−1,1]|∑ni=0fi(x)−∑∞i=0fi(x)| approach could be very helpful too!
Answer
The convergence is not uniform.
For x≠0 this is a geometric series:
∞∑n=0x2(1+x2)n=x2⋅11−11+x2=x2+1
and for x=0 the sum is 0.
Therefore
f(x)={x2+1, if x≠00, if x=0
Since f is not continuous, the convergence cannot be uniform.
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