# If $\sum_{n=1}^{\infty}|a_n| < \infty$. then $\sum_{n=1}^{\infty}f(a_nx)$ converges

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be differentiable at $0$ with $f(0) = 0$. Let $\{a_n\}$ be a real sequence such that $\sum_{n=1}^{\infty}|a_n| < \infty$. Show that $\sum_{n=1}^{\infty}f(a_nx)$ converges for every $x$ and that the sum of the series is differentiable at $0$.

I have included my attempt. Please tell me is that right way to do and what should be next; If not then what can be a better proof. I would welcome alternative proofs for this.

Since $$\sum_{k = 1}^\infty |a_k|<\infty$$ Given $0<\epsilon<1$, there exists $N$ such that for all $k\geq N$ $$\sum_{k = N}^\infty |a_k|<\epsilon$$ As a result, we have that $|a_k|<\epsilon$ for all $k\geq N$. Now, let $$M = \max\{|a_1|,|a_2|,\dots,|a_N|,\epsilon\}$$ Then, $|a_k|\leq M$ for all $k$. Now, choose $\delta$ such that for all $x$ with $|x|<\delta$: $$\left|\frac{f(n)}{n}-f'(0)\right|<\frac{\epsilon}{4M}$$ Then, choose $N_1$ such that $\forall n\geq N_1$, $|a_n n|<\delta$ (where $|n|<\frac{\delta}{4}$). We then choose $N_2$ such that $$\sum_{k = N_2}^\infty |a_k|<\frac{\epsilon}{4}$$ Then, we have that: $$\sum_{n = 1}^\infty \left(\frac{f(a_n n)}{a_n n}a_n-f'(0)a_n\right)\leq\sum_{n = 1}^{N' = \max\{N_1,N_2\}}\left|\frac{f(a_n n)}{a_n n}a_n-f'(0)a_n\right|+\sum_{n = N'+1}^{\infty}\left|\frac{f(a_n n)}{a_n n}-f'(0)\right||a_n|$$

• Mathjax please! – Hagen von Eitzen Jan 3 '17 at 7:48
• I've attempted to transcribe your argument, I'd appreciate you looking over it and mentioning any issues (I had some issues interpreting your handwriting). – Mark Jan 3 '17 at 8:05
• I assume $f(n)/n$ should be $f(x)/x$? – Clement C. Jan 3 '17 at 8:20
• Also, if you have access to theorems of comparison, the fact that $a_n x \xrightarrow[n\to\infty]{}0$ for any fixed $x\in\mathbb{R}$, and that therefore $$f(a_n x) =_{n\to\infty} f(0)+ f'(0) a_n x + o(a_n) = f'(0) a_n x + o(a_n)$$ would give a much more succinct argument. – Clement C. Jan 3 '17 at 8:23

Convergence is immediate, since the sequence of $a_n$ is bounded in absolute value by $M,$ since $|f(a_n x)| < 2 |a_n f^\prime(0)|,$ for $x$ small enough (where the existence of $M$ makes "small enough" uniform. The derivative of $f(a_n x)$ at $0$ is $a_n f^\prime(0),$ which gives that the derivative of the sum is the sum of the derivatives.
• @manhattan you use the fact that $f(x)-xf^\prime(0) = g(x) x,$ where $g(x)$ approaches $0$ at $0,$ and the speed of convergence is uniform (because of the $M,$ as above). – Igor Rivin Jan 3 '17 at 8:34
• I am not sure I buy that. If $f'(0)=0$, you are basically saying that $f$ is identically zero on a neighborhood of the origin. Am I misunderstanding? – Clement C. Jan 3 '17 at 9:02
• I think to apply comparison test u have to use the differentiability for some $1>\epsilon>0$ so that $f(a_nx) \leq a_n(f(0)+\epsilon)$ and appropriate a_ns – manhattan Jan 3 '17 at 9:08