# Thread: Limit of an integral sequence

1. ## Limit of an integral sequence

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.

2. Originally Posted by Mathventure
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 $I = \frac{\;\;nx^{n-1}}{1+x}$ and notice the following if we divide
$n = 1,\;\; I = \frac{1}{1+x}$
$n = 2,\;\; I = 2\left( 1 - \frac{1}{1+x} \right)$
$n = 3,\;\; I = 3\left( x - 1 + \frac{1}{1+x} \right)$
$n = 4,\;\; I = 4\left( x^2 - x + 1 - \frac{1}{1+x} \right)$
$n = 5,\;\; I = 5\left( x^3 - x^2 + x - 1 + \frac{1}{1+x} \right)$

3. Originally Posted by danny arrigo
Let $I = \frac{\;\;nx^{n-1}}{1+x}$ and notice the following if we divide
$n = 1,\;\; I = \frac{1}{1+x}$
$n = 2,\;\; I = 2\left( 1 - \frac{1}{1+x} \right)$
$n = 3,\;\; I = 3\left( x - 1 + \frac{1}{1+x} \right)$
$n = 4,\;\; I = 4\left( x^2 - x + 1 - \frac{1}{1+x} \right)$
$n = 5,\;\; I = 5\left( x^3 - x^2 + x - 1 + \frac{1}{1+x} \right)$
then what will be the exact answer and how....there r 4 options available
0,1/2,1,infinity

4. i got zero...if thats the answer, ill post the solution...

i got zero...if thats the answer, ill post the solution...
even i am nt sure of answer......whatever ur solution is, plz post it.

6. ## soln

here's my solution,

$\large S=\int_{0}^{1}\frac{nx^{n-1}}{1+x}$

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;

$\large S=n(\frac{x^{n}}{n(1+x)} + \int_{0}^{1}\frac{x^{n}}{n(1+x)^{2}})$

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,

$\large S=(\frac{x^{n}}{(1+x)} + \int_{0}^{1}\frac{x^{n}}{(1+x)^{2}})$

Now, applying the limits of integration in the first terms gives us,

$\large S=1/2 + \int_{0}^{1}\frac{x^{n}}{(1+x)^{2}}$

Then,

$\large \lim_{n\to inf }S=1/2 + \lim_{n\to inf }\int_{0}^{1}\frac{x^{n}}{(1+x)^{2}}$

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,

$\large \lim_{n\to inf }S=1/2$