If x∈[k,1] , 0<k<1 , then 1+n2x2x≤n2x2x=n2x1≤n2k1.
The expression on the right hand does not depend on x and the series n=1∑∞n2k1 is convergent, therefore the initial series is uniformly convergent on the segment [k,1].
In order to prove that the series is uniformly convergent for any k>0 , it is sufficient to show that for any N∈N there exists x∈[0,1] , q>p>N such that n=p∑q1+n2x2x≥1/5 .
Put x=1/N, p=N+1, q=2N. Then nx≤2 for all n=p,p+1,…,q and
n=p∑q1+n2x2x≥n=p∑q1+22x=n=N+1∑2N51/N=1/5
Therefore, the series n=1∑∞1+n2x2x is uniformly convergent for any k>0 .
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