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I am asked to show that a function $\sum \frac{1}{n^2} \cos (x)$ is differentiable on $\mathbb{R}$.

I have a theorem that states:

Let $I$ be a bounded interval, $x_0\in I$, and ${f_n}$ a sequence of functions such that $f_n$ is differentiable on $I$ for all $n\geq 1$ and the series $\sum f_n(x_0)$ converges and the series $\sum f'_n$ converges uniformly on $I$. Then the series $\sum f_n$ converges uniformly to some $f$ on $I$ such that $f$ is differentiable on $I$ and $f'=\sum f'_n$

I have verified all of the hypothesis but my question is how can I apply this theorem since this theorem only considers a bounded interval and I'm trying to show that it is differentiable on all of $\mathbb{R}$.

Thanks in advance.

  • 1
    What it means to be differentiable on $\mathbb{R}$? When can you say that a series is convergent (uniformly) on $\mathbb{R}$? – 2017-02-25

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To be differentiable on $\mathbb{R}$ means that for every $x \in \mathbb{R}$, the function is differentiable at $x$.

Pick any $x \in \mathbb{R}$ and a small interval $I$ containing $x$. Your theorem says that your function is differentiable on $I$, so it is differentiable at $x$. Thus it is differentiable everywhere.