Ξ£n=1ββx[(1+n2x2)β(1+(n+1)2x2)(n+1)βnβ]
We can reduce the series to a simpler form to work with as shown below
Ξ£n=1ββ[(1+nx)((nβ1)x+1)xβ]
The above series is uniformly convergent on any finite interval [a; b], when 0<a<b . But the series is only point wise convergent but not uniformly convergent on [a, b]
Here Unβ(x)=[(1+nx)((nβ1)x+1)xβ]=(nβ1)x+11ββnx+11β
Therefore snβ(x)=u1β(x)+u2β(x)+u3β(x)+β
β
β
+unβ(x)
snβ(x)=(1βx+11β)+(x+11β+2x+11β)+(2x+11β+3x+11β)+...+[(nβ1)x+11ββnx+11β]snβ(x)=1βnx+11ββ΄s(x)=Limnβββsnβ(x)=x={10βif xξ =0if x=0β
Therefore the sum function s(x) is discontinuous on [a, b], and therefore the convergence is not uniform
on [a, b], it is only point wise. When xξ =0 , let Ο΅>0 be given
β£snβ(x)βs(x)β£=β£1βnx+11ββ1β£=β£nx+11ββ£<Ο΅ ; whenever n>x1β(Ο΅1ββ1)
But x1β(Ο΅1ββ1) decreases with x. Hence if we take N=[xΟ΅[a,b]maxβ{x1β(Ο΅1ββ1)]+1Ο΅N
which is independent of x, then we obtain β£snβ(x)βs(x)β£<Ο΅ whenever n> N whenever n > N for all xΟ΅[a,b] i.e. the series converges uniformly to s(x) = 1 on [a,b].
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