$f$ such that $\sum a_n$ converges $\implies \sum f(a_n)$ converges
Solution 1:
I have added an answer to collect various resources where this fact is proven. This answer is community-wiki - feel free to add further references you are aware of.
Papers:
- Gerald Wildenberg: Convergence-Preserving Functions, The American Mathematical Monthly Vol. 95, No. 6 (Jun. - Jul., 1988), pp. 542-544, jstor link, doi: 10.2307/2322761, MR0945022, Zbl 0664.40001.
- Arthur Smith: Convergence-Preserving Functions: An Alternative Discussion, The American Mathematical Monthly Vol. 98, No. 9 (Nov., 1991), pp. 831-833, jstor link, doi: 10.2307/2324271, MR1133000, Zbl 0783.40002
- Rado, R. A theorem on infinite series. J. London Math. Soc. 35 1960 273–276; doi: 10.1112/jlms/s1-35.3.273, MR145321, Zbl 0098.26703
Books
- Teodora-Liliana Radulescu, Vicentiu D. Radulescu and Titu Andreescu: Problems in Real Analysis: Advanced Calculus on the Real Axis, Problem 2.5.30, page 108
Online resources
- The set of functions which map convergent series to convergent series
Solution 2:
A start: There exists $\delta > 0$ such that $\frac{f(x)}{x}$ is bounded in $\{0<|x|<\delta\}.$ Proof: Otherwise there is a sequence of nonzero terms $x_n \to 0$ such that
$$\left|\frac{f(x_n)}{x_n}\right|> n$$
for all $n.$ We can pass to a subsequence $n_k$ such that $|x_{n_k}| < 1/k^2$ for all $k.$ We then have
$$(1)\,\,\,\,\left|\frac{f(x_{n_k})}{x_{n_k}}\right|> n_k \ge k.$$
Consider now the case where $x_{n_k}, f(x_{n_k})> 0$ for all $k.$ Then for each $k,$ there is a positive integer $m_k$ such that
$$(2)\,\,\,\,\frac{1}{k^2} < m_kx_{n_k} < \frac{2}{k^2}.$$
Think about the series
$$x_{n_1} + x_{n_1} + \cdots + x_{n_1} + x_{n_2} + \cdots + x_{n_2} + x_{n_3} + \cdots ,$$
where $x_{n_1}$ occurs $m_1$ times, $x_{n_2}$ occurs $m_2$ times, etc. By $(2)$ this series converges. But note $(1)$ shows the $k$th grouping in the "$f$-series" sums to
$$m_kf(x_{n_k}) > m_k\cdot k\cdot x_{n_k} > \frac{1}{k}.$$
Thus the $f$-series diverges, contradiction. So we are done in the case $x_{n_k}, f(x_{n_k})> 0.$ The other cases follow by similar arguments..
Solution 3:
Here comes a proof in the general case. We first start with the following
Lemma: Let $-\varepsilon<y<0<x<\varepsilon$ and $N\in\mathbb{N}$. Then there is a some $M\in\mathbb{N}$ and $z_{1},\dots,z_{M}\in\left\{ x,y\right\} $ with $$ \left|\left\{ i\in\left\{ 1,\dots,M\right\} \,\mid\, z_{i}=x\right\} \right|=N $$ and $$ \left|\sum_{i=1}^{K}z_{i}\right|<\varepsilon\qquad\forall K\in\left\{ 1,\dots,M\right\} .\qquad\qquad\left(\ast\right) $$
Proof: We first inductively construct an infinite sequence $\left(w_{n}\right)_{n\in\mathbb{N}}\in\left\{ x,y\right\} ^{\mathbb{N}}$ with $\left|\sum_{i=1}^{n}w_{i}\right|<\varepsilon$ for all $n\in\mathbb{N}$. It is clear that the stated condition holds for $w_{1}:=x$, since $-\varepsilon<x<\varepsilon$. Now, let $w_{1},\dots,w_{n}$ be already constructed. Distinguish two cases:
We have $\alpha:=\sum_{i=1}^{n}w_{i}\geq0$. Then choose $w_{n+1}:=y$. Then $$ \varepsilon>\alpha>\alpha+y=\sum_{i=1}^{n+1}w_{i}=\alpha+y\geq y>-\varepsilon, $$ i.e. $\left|\sum_{i=1}^{n+1}w_{i}\right|<\varepsilon$.
We have $\alpha:=\sum_{i=1}^{n}w_{i}<0$. Then we choose $w_{n+1}:=x$. Similar to the above, it follows $$ -\varepsilon<\alpha<\alpha+x=\sum_{i=1}^{n+1}w_{i}=\alpha+x\leq x<\varepsilon $$ and hence $\left|\sum_{i=1}^{n+1}w_{i}\right|<\varepsilon$.
Now assume that $\left\{ i\in\mathbb{N}\,\mid\, w_{i}=x\right\} $ is finite. This implies $w_{i}=y$ for all $i\geq N_{0}$ with $N_{0}\in\mathbb{N}$ suitable. But this implies $$ \sum_{i=1}^{N}w_{i}=\sum_{i=1}^{N_{0}-1}w_{i}+\sum_{i=N_{0}}^{N}w_{i}=\sum_{i=1}^{N_{0}-1}w_{i}+\left(N-N_{0}\right)y\xrightarrow[N\to\infty]{}-\infty, $$ in contradiction to $\left|\sum_{i=1}^{N}w_{i}\right|<\varepsilon$ for all $N$.
Thus, $\left\{ i\in\mathbb{N}\,\mid\, w_{i}=x\right\} $ is infinite. It is easy to see that this implies existence of $M\in\mathbb{N}$ with $\left|\left\{ i\in\left\{ 1,\dots,M\right\} \,\mid\, w_{i}=x\right\} \right|=N$, so that we can set $z_{i}:=w_{i}$ for $i\in\left\{ 1,\dots,M\right\} $. $\square$
Remark: 1. A completely similar argument show that we can choose $z_{1},\dots,z_{M}\in\left\{ x,y\right\} $ with $$ \left|\left\{ i\in\left\{ 1,\dots,M\right\} \,\mid\, z_{i}=y\right\} \right|=N $$ and such that $\left(\ast\right)$ still holds.
- For the above sequence, we also have $$ \left|\sum_{i=K}^{M}z_{i}\right|=\left|\sum_{i=1}^{M}z_{i}-\sum_{i=1}^{K-1}z_{i}\right|\leq\left|\sum_{i=1}^{M}z_{i}\right|+\left|\sum_{i=1}^{K-1}z_{i}\right|<2\varepsilon\qquad\qquad\left(\square\right) $$ for all $K\in\left\{ 1,\dots,M\right\} $.
By the argument given by @zhw., there is some $\delta>0$ and $C>0$ with $\left|\frac{f\left(x\right)}{x}\right|\leq C$ for all $x\in\left(-\delta,\delta\right)\setminus\left\{ 0\right\} $. We will need this constant $C$ below.
First note that $f\left(0\right)=0$ is trivial (consider the sequence $a_{n}=0$). Now, let us assume towards a contradiction that $f$ is not linear on any neighborhood of $0$. This implies that for every $n\in\mathbb{N}$ there are $x_{n}\in\left(0,\frac{1}{n^{2}}\right)$ and $y_{n}\in\left(-\frac{1}{n^{2}},0\right)$ with $$ \alpha_{n}:=\frac{f\left(x_{n}\right)}{x_{n}}\neq\frac{f\left(y_{n}\right)}{y_{n}}=:\beta_{n}. $$ Indeed, if there where no such $x_{n},y_{n}$ for some $n\in\mathbb{N}$, we could fix some $y_{n}\in\left(-\frac{1}{n^{2}},0\right)$ and would get $$ \gamma:=\frac{f\left(y_{n}\right)}{y_{n}}=\frac{f\left(x\right)}{x} $$ for all $x\in\left(0,\frac{1}{n^{2}}\right)$. Likewise, fixing now an arbitrary $x_{n}\in\left(0,\frac{1}{n^{2}}\right)$ and get $$ \frac{f\left(y\right)}{y}=\frac{f\left(x_{n}\right)}{x_{n}}=\frac{f\left(y_{n}\right)}{y_{n}}=\gamma $$ for all $y\in\left(-\frac{1}{n^{2}},0\right)$. All in all, this implies $f\left(y\right)=\gamma\cdot y$ for all $y\in\left(-\frac{1}{n^{2}},\frac{1}{n^{2}}\right)$, since we have $f\left(0\right)=0$. Thus, $f$ is linear on a neighborhood of $0$ after all, a contradiction.
Note that (by choice of $C$ above), we have $\left|\alpha_{n}\right|\leq C$ and $\left|\beta_{n}\right|\leq C$. Let $\varepsilon_{n}:=\frac{1}{n^{2}}$ and choose $N_{n}\in\mathbb{N}$ with $N_{n}\geq\frac{1+C\varepsilon_{n}}{\left|\alpha_{n}-\beta_{n}\right|\left|x_{n}\right|}$. Note that this is possible, since $\left|\alpha_{n}-\beta_{n}\right|>0$ and $\left|x_{n}\right|>0$. By the Lemma above, we find some $M_{n}\in\mathbb{N}$ and $z_{1}^{\left(n\right)},\dots,z_{M_{n}}^{\left(n\right)}\in\left\{ x_{n},y_{n}\right\} $ with $$ N_{n}=\left|\left\{ i\in\left\{ 1,\dots,M_{n}\right\} \,\mid\, z_{i}^{\left(n\right)}=x\right\} \right| $$ and $$ \left|\sum_{i=1}^{K}z_{i}^{\left(n\right)}\right|<\varepsilon_{n}=\frac{1}{n^{2}}\qquad\forall K\in\left\{ 1,\dots,M_{n}\right\} .\qquad\qquad\left(\dagger\right) $$ In particular (for $K=M_{n}$), we get $$ \left|N_{n}x_{n}+\left(M_{n}-N_{n}\right)y_{n}\right|=\left|\sum_{i=1}^{M_{n}}z_{i}^{\left(n\right)}\right|<\frac{1}{n^{2}}.\qquad\qquad\left(\ddagger\right). $$
Now define a sequence $\left(a_{k}\right)_{k\in\mathbb{N}}$ by $$ a_{k}=z_{\ell}^{\left(K\right)}\qquad\text{ for }k=\sum_{i=1}^{K-1}M_{i}+\ell\text{ with }K\in\mathbb{N}\text{ and }\ell\in\left\{ 1,\dots,M_{K}\right\} . $$ It is not too hard to see that this is well-defined. Furthermore, the series $\sum_{k}a_{k}$ is convergent, since it is Cauchy; indeed, for $\sum_{i=1}^{T-1}M_{i}+p=t\geq k=\sum_{i=1}^{K-1}M_{i}+\ell$, we have \begin{eqnarray*} \left|\sum_{n=k}^{t}a_{n}\right| & \leq & \left|\sum_{n=\ell}^{M_{K}}z_{n}^{\left(K\right)}\right|+\sum_{S=K+1}^{T-1}\left|\sum_{n=1}^{M_{S}}z_{n}^{\left(S\right)}\right|+\left|\sum_{n=p}^{M_{T}}z_{n}^{\left(T\right)}\right|\\ & < & 2\varepsilon_{K}+\sum_{S=K+1}^{T-1}\varepsilon_{S}+\varepsilon_{T}\\ & \leq & 2\sum_{n=K}^{T}\frac{1}{n^{2}}\leq2\sum_{n=K}^{\infty}\frac{1}{n^{2}}\xrightarrow[K\to\infty]{}0. \end{eqnarray*}
By assumption on $f$, this implies convergence of $\sum_{n}f\left(a_{n}\right)$. In particular, this series is Cauchy. But for $K\in\mathbb{N}$, we have \begin{eqnarray*} \left|\sum_{n=1+\sum_{i=1}^{K-1}M_{i}}^{\sum_{i=1}^{K}M_{i}}f\left(a_{n}\right)\right| & = & \left|\sum_{n=1}^{M_{K}}f\left(z_{n}^{\left(K\right)}\right)\right|\\ & = & \left|N_{n}f\left(x_{n}\right)+\left(M_{n}-N_{n}\right)f\left(y_{n}\right)\right|\\ & = & \left|N_{n}\alpha_{n}x_{n}+\left(M_{n}-N_{n}\right)\beta_{n}y_{n}\right|\\ & = & \left|N_{n}\left(\alpha_{n}-\beta_{n}\right)x_{n}+\beta_{n}\left(N_{n}x_{n}+\left(M_{n}-N_{n}\right)y_{n}\right)\right|\\ & \geq & N_{n}\left|\alpha_{n}-\beta_{n}\right|\left|x_{n}\right|-\left|\beta_{n}\right|\cdot\left|N_{n}x_{n}+\left(M_{n}-N_{n}\right)y_{n}\right|\\ & \overset{\text{see }\left(\ddagger\right)}{\geq} & N_{n}\left|\alpha_{n}-\beta_{n}\right|\left|x_{n}\right|-C\varepsilon_{n}\\ & \geq & 1 \end{eqnarray*} by choice of $N_{n}$, which shows that $\sum_{n}f\left(a_{n}\right)$ is not Cauchy and hence not convergent, a contradiction.