# Show that if the series $\sum_{n=1}^\infty f_n$ is uniformly convergent on $\Bbb R$, then the sequence $(f_n)$ converges to $0$ on $\Bbb R$.

Let $$f_n:\Bbb R \to \Bbb R$$ be a function for which the series $$\sum_{n=1}^\infty f_n$$ is uniformly convergent on $$\Bbb R$$ for any $$n \in \Bbb N$$. Show that the sequence $$(f_n)$$ converges to $$0$$ on $$\Bbb R$$.

Definition. A sequence $$(f_n)$$ of functions on $$A \subseteq \Bbb R$$ to $$\Bbb R$$ converges to a function $$f:A \to \Bbb R$$ on $$A$$ if for any $$\epsilon>0$$ and each $$x \in A$$, there exists $$N \in \Bbb N$$ such that for all $$n \ge N$$, we have $$|f_n(x)-f(x)|<\epsilon$$.

Cauchy Criterion for Series of Functions. Let $$(f_n)$$ be a sequence of functions on $$A \subseteq \Bbb R$$ to $$\Bbb R$$. Then the series $$\sum_{n=1}^\infty f_n(x)$$ is uniformly convergent on $$A$$ iff for all $$\epsilon>0$$, there exists $$K \in \Bbb N$$ such that for any $$m>n\ge K$$, we have $$\left|\sum_{k=n+1}^m f_k(x) \right|<\epsilon,$$ for all $$x \in A$$.

Attempt: Since the series $$\sum_{n=1}^\infty f_n$$ is uniformly convergent on $$\Bbb R$$ for any $$n \in \Bbb N$$, then by the Cauchy Criterion for Series of Functions, for all $$\epsilon>0$$, there exists $$K \in \Bbb N$$ such that for any $$m>n \ge K$$, we have $$\left|\sum_{k=n+1}^m f_k(x) \right|<\epsilon,$$ for all $$x \in \Bbb R$$.

The goal is to show that $$f_n \to 0$$ on $$\Bbb R$$. Let $$\epsilon>0$$ and $$x \in \Bbb R$$ be arbitrary. Choose $$N=K \in \Bbb N$$ such that for all $$n \ge N$$, we have $$|f_n(x)-0|=|f_n(x)|<|f_{n+1}(x)+\ldots+f_m(x)|<\epsilon.$$ Hence, $$f_n \to 0$$ on $$\Bbb R$$.

Does this approach correct? Thanks in advanced.

Your use of the triangle inequality is wrong. You may think of partial sums. The proof is essentially identical to the proof of the property $$\sum_{n=1}^{\infty}a_n \text{ converges }\implies a_n\to 0$$

• So, what is the correct approach? How does it go? Jun 1, 2022 at 2:22
• I'll give you one more hint. The series converges if the partial sums $\sum_{k=1}^{n}f_k$ converges. Then $\sum_{k=1}^{n+1}f_k$ also converges as $n\to\infty$. What can you get from here? Jun 1, 2022 at 2:24

From Cauchy Criterion for Series of Functions (putting $$m:=n+1$$) one can get that for all $$\epsilon>0$$ there exists $$K \in \Bbb N$$ such that for any $$n\ge K$$, we have $$\left|f_{n+1}(x) \right|<\epsilon$$ for all $$x \in A$$. Therefore there exists $$K \in \Bbb N$$ (and equals 'previous' $$K$$ plus one) such that for any $$n\ge K$$, we have $$\left|f_{n}(x) \right|<\epsilon$$ for all $$x \in A$$. This shows that $$(f_n)$$ converges to zero uniformly.

Let $$f:\Bbb R \to \Bbb R$$ be a function such that $$\sum_{n=1}^\infty f_n(x)$$ converges uniformly on $$\Bbb R$$ to $$f$$. Then the series converges on $$\Bbb R$$ to $$f$$. Let $$s_n:=\sum_{k=1}^n f_k(x)$$ be the partial sum of this series. Notice that $$s_n(x)-s_{n-1}(x) = f_n(x)$$. Now, we have $$s_n \to f$$ and $$s_{n-1} \to f$$ on $$\Bbb R$$ as $$n \to \infty$$. By definition, for all $$\epsilon>0$$ and each $$x \in \Bbb R$$, there exists $$N_1 \in \Bbb N$$ such that for any $$n \ge N_1$$, we have $$|s_n(x)-f(x)| < \frac{\epsilon}{4}$$. Similarly, for all $$\epsilon>0$$ and each $$x \in \Bbb R$$, there exists $$N_2 \in \Bbb N$$ such that for any $$n \ge N_2$$, we have $$|s_{n-1}(x)-f(x)| < \frac{\epsilon}{4}$$. Let $$N:=\max\{N_1,N_2\}$$. Then $$N \in \Bbb N$$ and for all $$n\ge N$$, we have \begin{align*} |f_n(x)| &= |s_n(x)-s_{n-1}(x)| \\ &\le |s_n(x)-f(x)| + |s_{n-1}(x) - f(x)| \\ &< \frac{\epsilon}{4} + \frac{\epsilon}{4} \\ &= \frac{\epsilon}{2} \\ &< \epsilon. \end{align*} Thus, the sequence $$(f_n)$$ converges to $$0$$ on $$\Bbb R$$.