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, 100n=1,1.5,2,...,9.5,10,25,50,75,100. By then it was clearly reaching 0.50.5.

Apr 8, 2018

See below.

Explanation:

int_0^1 (n x^(n-1))/(1+x^2) dx le int_0^1n x^(n-1)dx = 110nxn11+x2dx10nxn1dx=1

int_0^1 (n x^(n-1))/(1+x^2) dx ge 1/2 int_0^1n x^(n-1)dx = 1/210nxn11+x2dx1210nxn1dx=12

or

1/2 le int_0^1 (n x^(n-1))/(1+x^2) dx le 11210nxn11+x2dx1

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

NOTE

for x in [0,1]x[0,1]
1/2 le 1/(1+x^2) le 11211+x21

Apr 9, 2018

1/212

Explanation:

As has already been shown in a previous solution,

I_n = int_0^1 (nx^(n-1))/(1+x^2)dxIn=10nxn11+x2dx

exists and is bounded :

1/2 le I_n <112In<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) dxIn=(nxn1dx1+x2)1010xn×(2x(1+x2)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