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Suppose $\sum_{n=1}^{\infty} a_n$ is a convergent series in $\mathbb{R}$. And, for all $a_n>0$ for all $n=1,2,3,...$.

Let $f%$ be a differential function, where $f(0)=0$.

I want to prove that $\sum_{n=1}^{\infty} f(a_n)$ also converges.

I'm thinking of using a Taylor series expansion for $f(x)$ to start my proof, so $$f(a_n)=f(0)+f'(0)a_n+o(1)a_n=a_n(f'(0)+o(1))$$ where $o$ is Peano's Remainder.

Is there a better way to start this?

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We have $|f(u)|\leq M|u|$ for $u\in(-\delta,\delta)$, $\delta>0$ is small. Now $|f(a_{n})|\leq Ma_{n}$ eventually, so $\displaystyle\sum_{n}|f(a_{n})|<\infty$.

Note that $a_{n}\rightarrow 0$. Also note that since $f$ is differentiable at zero, there is some $\delta>0$ such that $\left|\dfrac{f(u)-f(0)}{u-0}-f'(0)\right|<1$ for $u\in(-\delta,\delta)-\{0\}$, then $|f(u)|<(1+|f'(0)|)|u|$ for $u\in(-\delta,\delta)$.

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