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Let $f_n: [0,1] \to \mathbb{R}$, $f_n(x)=(-1)^n (1-x)x^n, \forall x\in [0,1], n \ge 0$. Prove that $\sum\limits_{n \ge 0} f_n(x)$ convergenes uniformly.

Let $S_n(x) = \sum\limits_{k = 0}^n f_k(x), \forall x \in [0,1], n \ge 0$.

Then,

$S_n(x)=1-x-x+x^2+x^2-x^3-x^3+x^4+...+(-1)^{n-1}x^{n-1}+(-1)^nx^n+(-1)^nx^n+(-1)^{n+1}x^{n+1}$

so

$S_n(x)=1-2x+2x^2-...+(-1)^n\cdot 2x^n+(-1)^{n+1}x^{n+1}, \forall x\in [0,1], n \ge 0$.

We know that $1+x^{2p+1}=(1+x)(x^{2p}-x^{2p-1}+x^{2p-2}-....+x^2-x+1)$, so, for $n=2p$, $p \ge 0$, we may write: $$ S_{2p}= 2(1-x+x^2-...+x^{2p})-1-x^{2p+1}=2 \cdot \frac{1+x^{2p+1}}{1+x}-1-x^{2p+1}$$ Now, it's easy to see that $\lim\limits_{p \to \infty}S_{2p}(x)=\frac{1-x}{1+x}, \forall x\in [0,1]$.

For $n=2p+1$, $p \ge 0$, we have: $$ S_{2p+1}(x)=2(1-x+x^2-...+x^{2p+2})-1-x^{2p+2}=2\cdot \frac{1+x^{2p+3}}{1+x} -1 -x^{2p+2} $$ In this case we have: $\lim\limits_{p \to \infty} S_{2p+1}(x) = \frac{1-x}{1+x}, \forall x\in [0,1]$.

So, $\lim\limits_{n \to \infty} S_n(x) =\frac{1-x}{1+x}, \forall x\in [0,1]$.

Now, it remains to prove that $\lim\limits_{n\to \infty} \sup\limits_{x \in [0,1]} |S_n(x)-\frac{1-x}{1+x}|=0$.

It's easy to see that $g_n(x)=|S_n(x)-\frac{1-x}{1+x}|=x^{n+1} \cdot \frac{1-x}{1+x}$.

How can I continue?

npatrat
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    A quick remark : The beginning seems too complicated. Just observe (using geometric series) that $S_n(x) = \sum_{k=0}^n (-1)^k(1-x)x^k = (1-x)\sum_{k=0}^n (-x)^k= (1-x)\frac{1-(-x)^{n+1}}{1+x}$. – user37238 Nov 13 '15 at 08:28
  • Note that $S_n(0)=1$ and $S_n(1)=0$ for all $n\in\mathbb N$. Also note that $|(-1)^k(1-x)x^k|\leq |x|^k$ for each $k$. Thus, for $x\in (0,1)$, the series converges absolutely. This may be helpful. – SamM Nov 13 '15 at 09:12
  • +1 for showing what you did. Starting from where you stopped, note that $g_n\to0$ uniformly if and only if $h_n\to0$, with $h_n(x)=x^{n+1}(1-x)$, compute the derivative $h'_n(x)=x^n(n+1-(n+2)x)$, deduce that $h_n$ is maximal at $x_n=\frac{n+1}{n+2}$ and that $\max h_n=h_n(x_n)\leqslant1-x_n=\frac1{n+2}\to0$, hence the convergence is indeed uniform. – Did Nov 13 '15 at 10:06

1 Answers1

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Hints : with your notation, $g_n'(x) = \frac{ x^{n-1}(-nx^2+n-2x)}{(x+1)^2}$. The sign of $g_n'(x)$ on $[0,1]$ is the same as the sign of $-nx^2+n-2x$ and consequently is non-negative on $\left[0, \frac{\sqrt{n^2+1}-1}{n}\right]$ and non-positive on $\left[\frac{\sqrt{n^2+1}-1}{n},1\right]$. Consequently, the maximum of $g_n$ on $[0,1]$ is reached at $x= \frac{\sqrt{n^2+1}-1}{n}$. A study of the convergence of $g_n\left( \frac{\sqrt{n^2+1}-1}{n} \right)$ should lead to a solution.

user37238
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    For a computationally slightly simpler version (to the point that the final answer becomes more or less obvious), see my comment on main. – Did Nov 13 '15 at 10:08