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Uniform Convergence of $nx^{2}(1-x)^{n}$ on $[0,1]$

My attempt:

criterion: suppose $f_n:I\to\ J$ is a sequence of functions which converges point wise to a function $f$, then the convergence is uniform if and only if $$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|=0$$.

I found out the point at which this function attains maximum. This happens at $x=2/(2+n)$. and the value of $f_{n}(x)$ at this point is $4/[2/n+1]$. So, with pointwise limit $f=0$, we have $$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|=0$$.

So, $f_{n}$ converges uniformly.

Is there a flaw in the proof? Please clarify. Also suggest any alternative proof which is more analytic rather than calculation based, I mean if bounds can be established so that I don't have to find the maxima in the interval.

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You found correctly the point at which $$ x \longmapsto f_n(x):=nx^{2}(1-x)^{n} $$ attains its maximum, that is $\displaystyle x=\frac{2}{n+2}$, but there is a mistake in your computation for the value of $f_n$ at this point, the right value is rather $$ \frac{4 n \left(1-\frac{2}{2+n}\right)^n}{(2+n)^2}\sim \frac{4}{e^2}\frac{1}{n}, \quad \text{as} \quad n \to +\infty, $$ and your function converges uniformly to $0$ on $[0,1]$.

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    $\begingroup$ Thanks for pointing that out. $\endgroup$ – Silver moon Feb 9 '15 at 8:31

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