Is #sum_(n->0)^oo g_n(x)# with #g_n(x) = arctan(\frac{2x}{x^2+n^3})# continuous ?

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Answer 1 · 2018-01-15T11:19:03

See below.

First the functional series converges for all #x in RR# because

calling #g_n(x) = arctan(\frac{2x}{x^2+n^3})# we have

#sum_(n=1)^oo max(g_n(x)) le M# because

#max g_n(x) = arctan(n^(-3/2)) < n^(-3/2)# and #sum_(n=1)^oo n^(-3/2) = zeta(3/2)# converges.

Second, we have #abs(g_n(x)-g_n(y)) le 2/n^3abs(x-y)# (Lipschitz continuity)

because for # x in [0,oo)#

#max(d/(dx)g_n(x)) = 2/n^3#

and then

#abs(f(x)-f(y)) le 2 (sum_(k=1)^oo k^-3) abs(x-y)# thus #f(x)# is continuous.

NOTE:

#sum_(k=1)^oo k^-3= zeta(3) = 1.2020569031595942...#