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I'm trying to show that if $u_n(x) = (1-x^2)^2 x^n$ then $\sum u_n$ is uniformly convergent on $[0,1]$.

The only thing I can think to use is Weierstraß M-test to show that $u_n < M_n$ where $\sum M_n$ converges, but I can't find a way to make the $M_n$ independent of $x$.

Ayman Hourieh
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Dave
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3 Answers3

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By studying the derivative of $u_n(x)$, we can find the maximum value on $[0,1]$. We have: $$ u'_n(x) = x^{n-1}(x^2 - 1)\left((n + 4)x^2 - n\right) $$

Thus, $u_n(x)$ reaches its maximum on $[0, 1]$ at: $$ m_n = \sqrt{\frac{n}{n+4}} $$

And we have: $$ u_n(m_n) = \left(\frac{n}{n+4}\right)^{n/2}\frac{16}{(n+4)^2} \le \frac{16}{n^2} = M_n $$

Since $\sum 1/n^2$ converges, it now follows that the convergence is uniform.

Ayman Hourieh
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Using AM-GM inequality, we get $$[u_n(x)]^2 = (1-x^2)^4 x^{2n}= 4^4.n^n.\underbrace{\frac{(1-x^2)}{4}...\frac{(1-x^2)}{4}}_{4 \quad times }.\underbrace{\frac{x^2}{n}...\frac{x^2}{n}}_{n \quad times } \le 4^4. n^n \left(\frac{1}{n+4} \right)^{n+4}=M_n:=\left( \frac{4}{n+4}\right)^4 \left( \frac{n}{n+4}\right)^n.$$

Chanh Le Van
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HINT: We can see the limit $u(x)=(1-x^2)^2\frac1{1-x}=(1+x)^2(1-x)$ from the geometric series. Use that the limit is continuous and all $u_n$ are $\ge 0$.

Hagen von Eitzen
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  • So if we call s_n = sum u_k, I've shown that this s_n tends to your u_x, and then I need to show that this |s_n - u_x| < epsilon, but then I'm faced with the problem of finding an n depending on epsilon but independent of x? – Dave May 19 '13 at 11:03