# Series $a_n$ converges implies series of function $f(a_n)$ converges

Let $$f : \mathbb{R} \to \mathbb{R}$$ has the following property:

For every series $$\sum_{n=1}^{\infty} a_n$$ converges implies $$\sum_{n=1}^{\infty} f(a_n)$$ converges.

Show that there exist $$M>0$$ and $$\epsilon>0$$

$$|f(x)|\leq M x$$ for every $$x\in \mathbb{R}, 0

I know that if $$\sum_{n=1}^{\infty} f(a_n)$$ is convergent then $$f(a_n)$$ is convergent as $$\lim f(a_n)=0$$. i don't have any idea, I need your helps

By contradiction, for any natural number $$n$$, take $$M=n^2$$ and $$\varepsilon=n^{-2}$$. Then there exists $$0 such that $$|f(x_n)|\geq 1$$. You can conclude.
• Contradict to fact that $|f(x_n)|<\epsilon$? – SutMar Apr 20 '19 at 14:55
• Contradict to the fact that $\sum x_n$ converges but $\sum f(x_n)$ doesn't – elidiot Apr 21 '19 at 15:02
Proof by contradiction. Let's assume that for every $$M$$ and $$\epsilon$$ there exist $$x\in[0,\epsilon]$$ such that $$|f(x)| \ge M x$$. In particular that means that for every $$n\in\mathbb{N}$$ there exist $$x_n\in[0,\frac{1}{n^2}]$$ such that $$|f(x_n)| \ge nx_n$$.
Let us divide sequence $$x_n$$ into two subsequences, one for which $$f(x_n)>0$$ and the other for which $$f(x_n)<0$$. One of these two subsewuences is guaranteed to be infinite; without a loss of generality we can assume it's the former, i.e. there exists an infinite growing sequence $$n(k)$$, $$k\in\mathbb{N}$$ such that $$0 < x_{n(k)} < \frac{1}{n(k)^2}$$, $$f(x_{n(k)})>n(k)x_{n(k)}$$. Obviously, it also satisfies $$n(k) \ge k$$.
Let us construct a seqence $$a_n$$ in the form $$x_{n(1)}, \dots x_{n(1)}, x_{n(2)}, \dots x_{n(2)}, x_{n(3)}, \dots x_{n(3)}, x_{n(4)}, \dots x_{n(4)}, x_{n(5)}, \dots$$ where number $$m_k$$ of copies of $$x_{n(k)}$$ is such that $$\frac{1}{k^2} < m_k x_{n(k)} < \frac{2}{k^2}$$ Since $$0 < x_{n(k)} < \frac{1}{n(k)^2} \le \frac{1}{k^2}$$, there's a guarantee that such $$m_k$$ exists. We have $$\sum_{n=1}^\infty a_n = \sum_{k=1}^\infty m_k x_{n(k)} < \sum_{k=1}^\infty \frac{2}{k^2}$$ so $$\sum a_n$$ is convergent. However, because $$f(x_{n(k)})> n(k)x_{n(k)} \ge kx_{n(k)}$$ we have $$\sum_{n=1}^\infty f(a_n) = \sum_{k=1}^\infty m_k f(x_{n(k)}) > \sum_{k=1}^\infty m_k k x_{n(k)} > \sum_{k=1}^\infty \frac{1}{k}$$ So $$\sum_n f(a_n)$$ is not convergent. The contradiction proves he theorem.