Sequence of functions - showing the series converges uniformly

Question

Consider the sequence of functions $$f_n:[0,1]\longrightarrow \mathbb{R},\qquad f_n:=(-1)^n(1-x)x^n,\qquad n\geq 0.$$

Show that the series $f(x):=\sum_{n=0}^{\infty}f_n(x)$ converges to $f$ uniformly on $[0,1]$.

I'm having trouble with the proof and looking for help, thanks :)

Answer 1

Using the summation formula of geometric sums, the partial sums are $$ \sum_{k=0}^nf_k(x)=\frac{1-x}{1+x}(1-(-x)^{n+1}). $$ Considering the cases $0\leq x<1$ and $x=1$ separately, we see that this converges pointwise to

$$ f(x)=\sum_{k=0}^{+\infty}f_k(x)=\frac{1-x}{1+x}\qquad\forall x\in[0,1]. $$

So the remainder is $$ R_n(x)=f(x)-\sum_{k=0}^nf_k(x)=\frac{1-x}{1+x}\left(1-(1-(-x)^{n+1})\right)=\frac{(1-x)(-x)^{n+1}}{1+x}. $$ An easy study of the derivative of the nonnegative function $g_n(x):=(1-x)x^{n+1}$ on $[0,1]$ shows that its maximum is attained for $x=\frac{n+1}{n+2}=1-\frac{1}{n+2}$. Hence $$ \sup_{[0,1]}\;|R_n(x)|\leq g_n\left(1-\frac{1}{n+2}\right)=\frac{1}{n+2}\left(1-\frac{1}{n+2}\right)^{n+1}\leq \frac{1}{n+2}. $$ It follows that $$ \lim_{n\rightarrow +\infty}\;\; \sup_{[0,1]}\;|R_n(x)|=0 $$ that is, the series converges uniformly to $f$ on $[0,1]$.

Answer 2

First, since $f_n(x)$ is alternating, and $f_n(x) \to 0$, the series $\sum_n f_n(x)$ converges, so $f(x)$ is well defined.

If $x \in[0,1)$ we have $\sum_{n=0}^\infty (-1)^n x^n = \frac{1}{1+x}$, and so we have $(1-x)\sum_{n=0}^\infty (-1)^n x^n = \frac{1-x}{1+x}$ for all $x \in [0,1]$. In particular, $|(1-x)\sum_{n=0}^\infty (-1)^n x^n| \le 1-x $ for all $x \in [0,1]$.

Now consider \begin{eqnarray} |f(x)-\sum_{n=0}^N (-1)^n (1-x)x^n| &=& |\sum_{n=N+1}^\infty (-1)^n (1-x)x^n |\\ &=& |\sum_{n=0}^\infty (-1)^{n+N+1} (1-x)x^{n+N+1} | \\ &=& x^{N+1} |\sum_{n=1}^\infty (-1)^{n} (1-x)x^{n} | \\ &\le& x^{N+1}(1-x) \\ & \le & \frac{1}{N+1}(\frac{N+1}{N+2})^{N+1} \\ & \le & \frac{1}{N} \end{eqnarray} It follows that the convergence is uniform.

Answer 3

Put $$F(x) = \sum f_n(x)\mbox{,}\qquad F_n(x)=\sum^n_{k=0}f_k(x)\mbox{.}$$ By Abel's theorem, since the series converges at $1$, $F$ is continuous at $1$.

Let $\epsilon>0$. Fix a positive even integer $n_0$. Observe that for $n\geq n_0$, $$F_{n_0}(x)\geq F_n(x)\geq0\qquad(0\leq x\leq1)\mbox{.}$$ Choose $1>\delta>0$ so that $$0\leq F(x)\leq F_{n_0}(x)<\epsilon/2$$ for $x\in[1-\delta,1]$. It follows that for $n\geq n_0$, $x\in[1-\delta,1]$, $$\begin{aligned} \lvert F_n(x)-F(x)\rvert&\leq \lvert F_n(x)-F_n(1)\rvert+\lvert F_n(1)-F(1)\rvert+\lvert F(1)-F(x)\rvert\\ &\leq F_{n_0}(x)+F(x)<\epsilon\mbox{,} \end{aligned}$$

By the Weierstrass M-test, $F_n\rightarrow F$ uniformly on $[0,1-\delta]$. Hence there exists an integer $N>n_0$ such that $n\geq N$ implies $$F_n(x)-F(x)\leq\epsilon$$ whenever $x\in[0,1-\delta]$. Consequently, $F_n\rightarrow F$ uniformly on $[0,1]$.