Prove that $\sum_{n=1}^{\infty}{f\left({1\over n}\right)}$ converges absolutely. Let $f:\Bbb{R}\to \Bbb{R}$ be continuously twice differentiable around $0$ such that $f(0)=f'(0)=0$. Prove that $\sum_{n=1}^{\infty}{f\left({1\over n}\right)}$ converges absolutely. 
What I did to far is:  $f(x)=o+o+R_n{(x,0)}={x^2 \over 2}f''(c)$. Therefore ${f\left({1\over n}\right)}={1 \over 2n^2}f''(c)$ where $0<c<{1\over n}$. By Heine definition of limits, $\lim_{n\to \infty}{f\left({1\over n}\right)}=f(0)=0$. I don't know how to show that either $f''(0)$ is bounded or uninfluential. I would really appreciate your help. 
 A: Let $c$ be so that $f''(x)$ is maximized on the interval $[0,1]$. Since $f''$ is continuous and the interval is compact, such a $c$ will exist. Then observe that
$$
|f(x)| \leq \frac{x^2}{2} |f''(c)|
$$
for any $x$ in $[0,1]$. This is due to your observation that by the taylor expansion, we have that $f(x) = x^2 f''(c')/2$ for some $c'$.
Then
$$
\sum_{n=1}^{\infty}\left|f(\frac{1}{n})\right| \leq \sum_{n=1}^{\infty} \frac{\left|f''(c) \right|
}{2n^2} = \frac{\left|f''(c) \right|
}{2} \sum_{n=1}^{\infty} \frac{1}{n^2}.
$$
The latter series converges to $\pi^2/6$. However, we do not need this fact, but only that either by the integral test or ratio test or whatever your favorite convergence test is, it converges. Therefore the sum in question converges absolutely, as desired.
A: Taylor's theorem with Peano form of the remainder tells us that $f(x)=f''(0)\frac{x^2}{2}+o(x^2)$
Hence $f(\frac{1}{n})=f''(0)\frac{1}{2n^2}+o(\frac{1}{n^2})$.
This also means that $\displaystyle\lim_{n\to \infty} \frac{f(\frac{1}{n})}{f''(0)\frac{1}{2n^2}}=1$
Therefore, there is some $N$ such that $\forall n\geq N$, $f(\frac{1}{n})$ and $f''(0)\frac{1}{2n^2}$have the same sign.
Thus, we may assume without loss of generality that $\forall n\in \mathbb N ,f(\frac{1}{n})\geq 0$

It remains to prove that $\sum f(\frac{1}{n})$ converges.
The statement $f(\frac{1}{n})=f''(0)\frac{1}{2n^2}+o(\frac{1}{n^2})$ may be rewritten as follows : 
there exists a sequence $\epsilon_n$ such that $\displaystyle\lim_{n\to\infty}\epsilon_n=0$ and $\forall n\in \mathbb N, f(\frac{1}{n})=f''(0)\frac{1}{2n^2}+\epsilon_n\frac{1}{n^2}$
Using $p$ series, $\sum_{n\geq 1} f''(0)\frac{1}{2n^2}$ converges.
Now, since $\displaystyle\lim_{n\to\infty}\epsilon_n=0$, there is some $N$ such that $\forall n\geq N,|\epsilon_n|\leq 1 $.
By comparison test, $\sum_{n\geq N}\epsilon_n\frac{1}{n^2}$ converges.
Hence $\sum_{n\geq 1}\epsilon_n\frac{1}{n^2}$ converges, and $\sum_{n\geq 1} f''(0)\frac{1}{2n^2}+\epsilon_n\frac{1}{n^2}$ converges too.
Hence $\sum f(\frac{1}{n})$ converges.
A: Since there are no answers yet, I thought I might post one and get it criticized:
Proof: $f({1\over n})={1\over 2n^2}f''(c)$. Since $0<c<{1\over n}$ then $c\to 0$ as $n\to \infty$ which means that there exists $m$ such that $f''(x)$ is continuous on $0\le x \le {1\over m} $, i.e, continuous on the closed interval $[0,{1\over m}]$ and therefore it attains its absolute maximum, $s$. Since $c\in [0,{1\over m}]$ then $|f''(c)|\le s$ from $m$ and on. Let us look at the tail $\sum_{i=m}^{\infty}{1\over 2n^2}|f"(c)|\le s\cdot \sum_{i=m}^{\infty}{1\over 2n^2}$ and since $\sum_{i=m}^{\infty}{1\over 2n^2}$ converges then so does $s\cdot \sum_{i=m}^{\infty}{1\over 2n^2}$ and so the tail of the original series, meaning, the series absolutely converges. 
