Prove $\sum\left| f\left(\frac 1 n\right)\right|^r$ converges for all $r>1$. I just got out of an introductory real analysis exam. I had this question which I tried to solve for more than $1.5$ hours and yet I accomplished nothing.


  Let $f:\Bbb R \to\Bbb R$ be a differentiable function with $f(0)=f'(0)=0$.
  
  Prove that 
  $$
\sum_{n=1}^\infty \left|f\left(\frac 1 n \right)\right|^r
$$
  Converges for all $r&gt;1$.


I tried to use the ratio test, root test, Lagrange's theorem, and a few more things I know but I couldn't crack this. 

Could someone explain how to solve this? (We only studied basic things about series)
Since $f$ is differentiable at $0$, the expression
$$
{f(\frac1n)-f(0)\over\frac1n}
$$
tends to $f'(0)=0$ as $n\to\infty$. But $f(0)=0$, so this means
$
nf(\frac1n)
$
tends to zero. In particular for all large $n$ we have
$$
\textstyle\left|nf(\frac1n)\right|\le1.
$$
Rearranging, this implies
$$
\left|f({\textstyle\frac1n})\right|^r\le\frac1{n^r}
$$
for all large $n$, which is enough to show $\sum|f(\frac1n)|^r$ converges.
By the MVT, $f(\frac{1}{n})=\frac{1}{n}\frac{f(\frac{1}{n})-f(0)}{\frac{1}{n}}=\frac{1}{n}f^\prime(x_n)$ with $0&lt;x_n&lt;\frac{1}{n}$. Therefore $x_n\to 0$. Then $|f(\frac{1}{n})|=\frac{1}{n}|f^\prime(x_n)|$.  Thus $f^\prime(x_n)\to f^\prime(0)=0$ since $nf(\frac{1}{n})=f^\prime(x_n)$ and $f$ is continuous. It implies that $\{f^\prime(x_n)\}$ is bounded. Say $|f^\prime(x_n)|\le M$.

Therefore $\sum_{n=1}^\infty|f(\frac{1}{n})|^r\le \sum_{n=1}^\infty|\frac{1}{n}f^\prime(x_n)|^r\le\sum_{n=1}^\infty\frac{1}{n^r}M^r=M^r\sum_{n=1}^\infty\frac{1}{n^r}&lt;\infty$