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Given $f_n(x)=(n+1)x^n; x\in [0,1]$

I want to show $\int_{[0,1]}f<\liminf\int_{[0,1]}f_n$, where $f_n$ converges pointwise to $f$ almost everywhere on $[0,1]$.

I have found that $\liminf\int f_n = \int f +\liminf\int|f-f_n|$, but I'm not sure how to use this, and I don't even know what $f_n$ converges to here. Can someone hint me in the right direction?

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1 Answer 1

HINT

Consider $a \in [0,1)$. The main crux is to compute $$\lim_{n \to \infty} (n+1)a^n$$ To compute the limit note that $a = \dfrac1{1+b}$ for some $b > 0$.

Hence, $$a^n = \dfrac1{(1+b)^n} < \dfrac1{\dfrac{n(n-1)}2 b^2}\,\,\,\,\,\,\,\,\, \text{(Why? Hint: Binomial theorem)}$$ Can you now compute $\lim_{n \to \infty} (n+1)a^n$?

HINT: $$\lim_{n \to \infty} (n+1)a^n < \lim_{n \to \infty} \dfrac{2n+2}{n(n-1) b^2} = ?$$

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zero! thanks for the hint. –  D. Rod Nov 5 '12 at 5:21

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