Proof of this series converging to zero does not convince me I have a function $F$ that is bounded and such that $F(x) \to 0$ whenever $x\to 0$. Then, I have the following series
$$
\sum_{k=0}^\infty \frac{F(2^{k+1}\delta)}{2^k}
$$
for $\delta > 0$. I want to show that this series converges to $0$ as $\delta\to 0$.
Proof from the book which I don't understand
Given $\epsilon > 0$ we choose $N$ so large that
$$
\sum_{k \geq N} \frac{1}{2^k} < \epsilon.
$$
Then, by making $\delta$ sufficiently small, we have (since $F(x)\to 0$ as $x\to 0$)
$$
F(2^k \delta) < \frac{\epsilon}{N}
$$
whenever $k=0, 1, \ldots, N-1$. Since $F$ is bounded then the result follows.
What I don't understand
I am very lost by this proof. What is it doing in the first step? How does it know I can find such an $N$? And how does it know that by choosing $\delta$ sufficiently small, I can bound $F(2^k \delta)$?
 A: *

*In the first step, they take advantage of the fact that $F$ is bounded and $\sum_{k\in\Bbb N}\frac1{2^k}$ is convergent to prove that some tail of the series is sufficiently small, independently from $\delta.$ More precisely, given $\epsilon>0,$ for $N$ large enough (and this "large enough" is easy to calculate),
$$\sum_{k\ge N}\frac1{2^k}=\frac1{2^{N-1}}<\epsilon$$
hence if $|F|$ is bounded by some positive $M$ then $\left|\sum_{k\ge N}\frac{F(2^{k+1}\delta)}{2^k}\right|\le\sum_{k\ge N}\frac{|F(2^{k+1}\delta)|}{2^k}\le M\epsilon$ for every $\delta.$

*Now that $N$ is fixed, in the second step, there only remains to bound a finite number of terms of the series. For each $k<N,$ they prove that for every $\delta$ small enough, each of the $N$ first terms of the series will also be small. More precisely, given $\epsilon>0$ again, since $\lim_{x\to0}F(x)=0,$ there exists some $\alpha>0$ such that
$$|x|<\alpha\Rightarrow|F(x)|<\frac\epsilon N,$$
so that for every real number $\delta,$
$$|\delta|<\frac\alpha{
2^{N-1}}\implies\forall k\le N-1\quad|F(2^k\delta)|<\frac{\epsilon}N$$
hence
$$|2\delta|<\frac\alpha{
2^{N-1}}\implies\sum_{k=0}^{N-1}\frac{|F(2^{k+1}\delta)|}{2^k}\le \sum_{k=0}^{N-1}|F(2^{k+1}\delta)|<\epsilon.$$
Putting the two steps together, since
$$\left|\sum_{k\ge0}\frac{F(2^{k+1}\delta)}{2^k}\right|\le\sum_{k=0}^{N-1}\frac{|F(2^{k+1}\delta)|}{2^k}+\left|\sum_{k\ge N}\frac{F(2^{k+1}\delta)}{2^k}\right|,$$
they found that
$$\forall\epsilon>0\quad\exists\beta>0\quad\left(|\delta|<\beta\implies\left|\sum_{k\ge0}\frac{F(2^{k+1}\delta)}{2^k}\right|<\left(1+M\right)\epsilon\right).$$
For every $\varepsilon>0,$ applying this to $\epsilon=\frac\varepsilon{1+M}$ proves the claim, i.e.
$$\forall\varepsilon>0\quad\exists\beta>0\quad\left(|\delta|<\beta\implies\left|\sum_{k\ge0}\frac{F(2^{k+1}\delta)}{2^k}\right|<\varepsilon\right).$$
A: 
What is it doing in the first step? How does it know I can find such an $N$?

Because $\sum_{k\geq N}\frac1{2^k} = \frac1{2^N}\sum_{k=0}^\infty\frac{1}{2^k}=2^{-N}2=2^{1-N}$ and we just choose $N\geq1-\log_2(\varepsilon)$.

And how does it know that by choosing $\delta$ sufficiently small, I can bound $F(2^k\delta)$?

The precise definition of $F(x)\to0$ as $x\to 0$ is:
$$\forall\varepsilon>0\exists\mu\forall x\colon |x|<\mu\Rightarrow |F(x)|<\varepsilon$$
(The usual definition uses $\delta$ instead of $\mu$, but we already have a different $\delta$ so I use $\mu$ to avoid confusion).
This means that if I choose some $\varepsilon$, for example $\varepsilon=\frac{\epsilon}{N}$, then I get a $\mu$ such that $F(x)<\varepsilon$ for all $x$ with $|x|<\mu$. Now if we take $\delta = \frac{\mu}{2^{N}}$, then all numbers $\delta 2^k$ for $k=1,2,\dots,N-1$ have $\delta 2^k<\mu$, so we have $F(\delta 2^k)<\varepsilon=\frac\epsilon N$.
A: the idea is to split the series into two separate sums. That means:
$\big\vert \sum_{k \geq 0} \frac{F(2^{k+1}\delta)}{2^k} \big\vert \leq \big\vert \sum_{k=0}^{N-1} \frac{F(2^{k+1}\delta)}{2^k} \big\vert + \big\vert \sum_{k\geq N} \frac{F(2^{k+1}\delta)}{2^k} \big\vert.$
Now, both sums on the right hand side tend to zero, under different assumptions. The first one, by using the continuity of $F$ in $0$, combined with $F(0)=0$ and the second one, using the boundedness of $F$. In fact you can proceed the inequality as follows.
For the first term:
$\big\vert \sum_{k=0}^{N-1} \frac{F(2^{k+1}\delta)}{2^k} \big\vert \leq N \max_{k=0,...,N-1} \vert F(2^{k+1} \delta) \vert. \tag{1}$
And the second term:
$\big\vert \sum_{k\geq N} \frac{F(2^{k+1}\delta)}{2^k} \big\vert \leq M \sum_{k\geq N} \frac{1}{2^k}. \tag{2}$
Note, such an $M>0$ exists, since $F$ is bounded.
Now you are ready to prove the claim. You can pick an arbitrary $\varepsilon >0$. Then, since the second sum (2) tends to zero, you find some $N(\varepsilon) \in \mathbb{N}$ s.t. $\sum_{k\geq N(\varepsilon)} \frac{1}{2^k} < \frac{\varepsilon}{2}$.
And, you can also find an $\delta(\varepsilon) >0$ small enough, s.t.
$\vert F(2^{k+1} \delta) \vert < \frac{\varepsilon}{2 \cdot N(\varepsilon)}, \quad \forall \delta \leq \delta(\varepsilon)$, for all $k=0,...,N(\varepsilon)-1$. Which gives you
$N(\varepsilon) \max_{k=0,...,N(\varepsilon)-1} \vert F(2^{k+1} \delta) \vert \leq \frac{\varepsilon}{2}, \quad \forall \delta \leq \delta(\varepsilon).$
Put everything together, and you get
$\big\vert \sum_{k \geq 0} \frac{F(2^{k+1}\delta)}{2^k} \big\vert \leq \varepsilon, \; \forall \delta \leq \delta(\varepsilon).$
This proves the claim, as $\varepsilon$ was arbitrary.
A: Let $\sum_{k=0}^{\infty}a_k = S \in \mathbb{R}$ be a convergent series. Now define a "tail sum" sequence: $t_n = \sum_{k=n}^{\infty}a_k$ 
Notice that this sequence is well defined because the series converges. Now we will prove that: $t_n \underset{n \to \infty}{\longrightarrow} 0$ 
Notice that $t_{n+1}=S-\sum_{k=0}^{n}a_k$ whoever we know that by definition $\sum_{k=0}^{n}a_k \underset{n \to \infty}{\longrightarrow} S$ so using arithmetic of limits we get $t_{n+1} \underset{n \to \infty}{\longrightarrow} S-S=0$ and thus also $t_n \underset{n \to \infty}{\longrightarrow} 0$ (because shifting a sequence does not impact the convergence or the limit itself).  Now substitute $a_k = 2^{-k}$ and use the definition of a limit of sequence on $t_n$ to get your desired $N$ for each $\epsilon$. 
Regarding your second doubt about $\delta$ consider the following: 
Let $0 < \epsilon, N$ be two numbers, we know that by the definition of the limit on the number $\epsilon/N$ there is some $0 < d$ such that $x \in (-d,d) \implies -\epsilon/N < F(x) < \epsilon/N$. Now the question is how to choose $0 < \delta$ such that $2^k\delta \in (-d,d)$ for $k=0,...,N-1$. Now note: 
$-d < 2^k\delta < d \Leftrightarrow -d/2^k < \delta < d/2^k$ so pick some $0<\delta$ such that: $-d/2^{N-1} < \delta < d/2^{N-1}$ and then because $(-d/2^{N-1},d/2^{N-1}) \subset (-d/2^k,d/2^k)$ for $k = 0,...,N-1$ that $\delta$ satisfies our requirement.
