how to solve
If S= integral of n*x^(n-1)/(1+x) for n>=1 with limit 0 to 1
then find limit of sequence {S} if n tends to infinity.
Let $\displaystyle I = \frac{\;\;nx^{n-1}}{1+x}$ and notice the following if we divide
$\displaystyle n = 1,\;\; I = \frac{1}{1+x}$
$\displaystyle n = 2,\;\; I = 2\left( 1 - \frac{1}{1+x} \right)$
$\displaystyle n = 3,\;\; I = 3\left( x - 1 + \frac{1}{1+x} \right)$
$\displaystyle n = 4,\;\; I = 4\left( x^2 - x + 1 - \frac{1}{1+x} \right)$
$\displaystyle n = 5,\;\; I = 5\left( x^3 - x^2 + x - 1 + \frac{1}{1+x} \right)$
i had made a mistake. its 1/2
here's my solution,
Now, our aim is to get rid of the n in the first place...
i used by parts for that.
using by parts taking 1/(1+x) as the first function, we get;
The first term of the sum has the limits 0->1 & we get rid of n from both terms by multiplying the n outside...
So,
Now, applying the limits of integration in the first terms gives us,
Then,
Now, observe the second term, here, x lies between 0 & 1. so x is basically a fraction & if we increase the power of a fraction, it decreses.
If we take the power to infinity, it almost (tends to) 0.
Hence, we have,