Recurrence relation proof

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November 11th 2011, 03:45 AM
Shizaru

Recurrence relation proof

$n = 1,2,3...$

$I_{n} = \int \frac {x^{n-1}}{2-x} dx$ between limits 0 and 1.

Writing $x^{n} = x^{n-1}(2-(2-x))$ , show this sequence satisfies $I_{n+1} = 2I_{n} - \frac {1}{n}$ .

Now, I can't quite get a foothold into this puzzle. I'm thinking there's a simple logical step but I don't see it. Can someone help please? I wanted to try induction but I don't see how I'm supposed to use the idea they've told me to do so. It is required of the problem to use this hint.
November 11th 2011, 04:44 AM
Danny

Re: Recurrence relation proof

Use the hint

$I_{n+1} = \int_0^1 \frac{x^n}{2-x}\,dx = \int_0^1 \frac{x^{n-1}\left(2 - (2-x)\right)}{2-x}\,dx = \int_0^1 \frac{2x^{n-1}}{2-x}\,dx - \int_0^1 x^{n-1} \frac{2-x}{2-x}\,dx$

so

$I_{n+1} = 2 I_{n} - \int_0^1 x^{n-1}\,dx$ .

Now integrate the last term.

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