There is a function which is continuous but not differentiable I have a function which is a convergent series:
$$f(x) = \sin(x) + \frac{1}{10}\sin(10x) + \frac{1}{100}\sin(100x) + \cdots \frac{1}{10^n}\sin(10^nx)$$
This function is convergent because for any E you care to specify, the function has a term which is smaller than E.  However, the function is not differentiable, and I don't understand why.
$$\frac{d}{dx}f(x) = \cos(x)+\frac{10}{10}\cos(10x) + \frac{100}{100}\cos(100x) + \cdots \frac{10^n}{10^n}\cos(10^nx)$$
Is this a special case to A Continuous Nowhere-Differentiable Function ?
 A: First note that $\sum a_k$ is absolutely convergent if 
$$\sum \left|a_k\right|\Rightarrow  \mbox{converges}$$
Since absolute convergence implies convergence, then let's check for absolute convergence 
$$ \sum\limits_{k=0}^n \left|\frac{\sin\left(10^kx\right)}{10^k}\right|$$
First note that
$$ \left|\sin\left(10^kx\right)\right|\leq 1 $$
$$ \frac{\left|\sin\left(10^kx\right)\right|}{\left|10^k\right|} \leq \frac{1}{\left|10^k\right|} $$
$$ \left|\frac{\sin\left(10^kx\right)}{10^k}\right| \leq \left|\frac{1}{10}\right|^k $$
Since $\left|\frac{1}{10}\right|\lt 1$, then by the geometric series test,
$$ \sum\limits_{k=0}^n \left|\frac{1}{10}\right|^k \Rightarrow \mbox{converges} $$
Which implies that
$$ \sum\limits_{k=0}^n \left(\frac{1}{10}\right)^k \Rightarrow \mbox{converges absolutely} $$
Therefore by the direct comparison test, we have
$$ f(x)=\sum\limits_{k=0}^n \frac{\sin\left(10^kx\right)}{10^k} \Rightarrow \mbox{converges absolutely}$$
Now let's talk about the differentiability of $f(x)$. If we take the derivative of each term in the sum, we'll end up with
$$ \sum\limits_{k=0}^n \cos\left(10^kx\right) $$
Now let's show that this series diverges. Note that 
$$ -1\leq \cos\left(10^nx\right) $$
Since 
$$ \lim\limits_{n\to\infty} 1=1\not=0 $$
Then by the divergence test
$$ -\sum\limits_{k=0}^n 1\Rightarrow \mbox{diverges} $$
Therefore by the direct comparison test
$$ \sum\limits_{k=0}^n \cos\left(10^kx\right) \Rightarrow \mbox{diverges}$$
So although each term of $f(x)$ is differentiable and the sum of each term is convergent, the sum of the derivative of each term does not converge. This fact makes $f(x)$ a non differentiable function. 
A: Notice that if you plug $x=k\pi$, where $k$ is any integer, into the derivative of the that function, you get $\infty$, thereby making the derivative discontinuous. This, in other words, means that the function is not differentiable.
