Sequence of functions $(f_n)$ with $|f_{n+1}(x)-f_n(x)| \le \frac{1}{2^n+x^2}$ is uniformly convergent in $\Bbb R$

Question

Let $(f_n)$ be sequence of functions with $f_n : \Bbb R \to \Bbb R$ satisfying $$|f_{n+1}(x)-f_n(x)| \le \frac{1}{2^n+x^2}, \quad \forall x \in \Bbb R, \forall n \in \Bbb N.$$ Show that $(f_n)$ is uniformly convergent in $\Bbb R$.

How to approach this? I didn't know yet how to start to solve this problem. Any ideas?

Answer 1

Cauchy Criterion for Uniform Convergence of Sequence of Functions:

A sequence of functions $(f_n)$ defined on a set $A\subseteq\mathbb{R}$ converges uniformly on $A$ if and only if for every $\epsilon>0$ there exists an $N\in\mathbb{N}$ such that $|f_m(x)-f_n(x)|<\epsilon$ whenever $m, n\geq N$ and $x\in A$.

Observation:

1. For all $x\in \Bbb R, x^2 \ge 0$.
2. For all $n\in \Bbb N, n < 2^n$.
3. If $n \in \Bbb N$, then $n+1 \in \Bbb N$. Further, if $n \ge N$, then $n+1 \ge N+1 \ge N$. Define $m:=n+1, \forall n \in \Bbb N$.

Let $\epsilon>0$ be given. By the Archimedean Principle, there exists $N \in \Bbb N$ with $\frac{1}{N} < \epsilon$ such that for any $m,n \ge N$, we have \begin{align*} |f_m(x)-f_n(x)| &= |f_{n+1}(x) - f_n(x)| \\ &\le \frac{1}{2^n+x^2}\\ & < \frac{1}{2^n} \\ &< \frac{1}{n} \\ &\le \frac{1}{N} \\ & < \epsilon, \end{align*} for all $x \in \Bbb R$.

Hence, by the Cauchy Criterion for Uniform Convergence of Sequence of Functions above, we have $(f_n)$ is uniformly convergent on $\Bbb R$. Q. E. D.

Answer 2

Hint: for $k\geq 1$, $$ f_{n+k}(x)-f_n(x)=\sum_{j=n}^{n+k}f_{j+1}(x)-f_j(x). $$

edit: since there seems to be some confusion in this post, I'll edit this to make a complete answer. The idea is to use the telescoping above, and the summable estimate on successive terms: for any $k\geq 1$, we have

$$ |f_{n+k}(x)-f_n(x)|=\left|\sum_{j=n}^{n+k}f_{j+1}(x)-f_j(x)\right|\\ \leq \sum_{j=n}^{n+k}|f_{j+1}(x)-f_j(x)|\\ \leq \frac{2^k-1}{2^{n+k-1}}\to 0 $$ as $n\to \infty$.

Note: the accepted takes a $m=n+1$, and claims the Cauchy criterion is satisfied, which of course is a statement for arbitrary $m$ and $n$. Note that by taking $$ f_{n}\equiv \sum_{k=1}^n\frac{1}{k}, $$ this proof would claim convergence of the harmonic series.