Kindly solve this? which option is correct?

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3 Answers
Mar 31, 2018

This is readily seen as not doable by elementary means, so I just solved it numerically and got:

I evaluated the integral for #n = 1, 1.5, 2, . . . , 9.5, 10, 25, 50, 75, 100#. By then it was clearly reaching #0.5#.

Apr 8, 2018

Answer:

See below.

Explanation:

#int_0^1 (n x^(n-1))/(1+x^2) dx le int_0^1n x^(n-1)dx = 1#

#int_0^1 (n x^(n-1))/(1+x^2) dx ge 1/2 int_0^1n x^(n-1)dx = 1/2#

or

#1/2 le int_0^1 (n x^(n-1))/(1+x^2) dx le 1#

Now assuming that one of the answers is true, the most natural seems to be the fourth 4)

NOTE

for #x in [0,1]#
#1/2 le 1/(1+x^2) le 1#

Apr 9, 2018

Answer:

#1/2#

Explanation:

As has already been shown in a previous solution,

#I_n = int_0^1 (nx^(n-1))/(1+x^2)dx#

exists and is bounded :

#1/2 le I_n <1#

Now integration by parts yields

# I_n = ((int nx^(n-1) dx)/(1+x^2))_0^1-int_0^1 x^n times (-(2x)/(1+x^2)^2) dx#
#qquad = (x^n/(1+x^2))_0^1+2int_0^1 x^(n+1)/(1+x^2)^2dx#
#qquad = 1/2 +J_n#

Now , since # 0 < (1+x^2)^-1 < 1# in #(0,1)#

#J_n = 2/(n+2) int_0^1 ((n+2)x^(n+1))/(1+x^2)^2 dx#
#qquad <= 2/(n+2) int_0^1 ((n+2)x^(n+1))/(1+x^2) dx = 2/(n+2)I_(n+2)#

Since #lim_(n to oo)I_n# exists, we have

#lim_(n to oo )J_n = lim_(n to oo) 2/(n+2)I_(n+2) = lim_(n to oo)2/(n+2) times lim_(n to oo)I_(n+2) = 0#

Hence

# lim_(n to oo) I_n = 1/2#