proving or disproving that $\sum_{k=0}^{N-1} {N-1\choose k}\epsilon^k - 1 < 1.01 N\epsilon$ 
Prove or disprove the following: if N is a positive integer and $\epsilon > 0$ is such that $N < \frac{.01}{\epsilon}$, then $\sum_{k=0}^{N-1} {N-1\choose k}\epsilon^k - 1 < 1.01 N\epsilon$.

Clearly from the question, we must have $\epsilon <.01$. Using the binomial theorem, we need to prove $(1+\epsilon)^{N - 1} - 1 < 1.01 N\epsilon$. I tried to approximate this using Taylor's theorem, which states that $(1+\epsilon)^{N-1} = 1  + (N - 1)\epsilon + {N-1\choose 2} (1+x)\epsilon^2,$ where $x$ is between $0$ and $\epsilon$, but I'm not sure if this is good enough. Perhaps it might be useful to prove a stronger/weaker inequality?

Below is my attempt to use the mean-value theorem for the function $f(x) = (1+x)^{N-1}$ on the interval $[0,\epsilon]$. The derivative of this function is $(N-1)(1+x)^{N-2}$. By the mean value theorem, there exists some $c$ in $(0,\epsilon)$ so that $(N-1)(1+c)^{N-2} = \frac{(1+\epsilon)^{N-1} - 1}{\epsilon},$ so we need to show $(N-1)(1+c)^{N-2} \epsilon < 1.01 N\epsilon^2$. From above, $\epsilon < 0.01$, so $c < 0.01$. For $N = 1,2,$ the inequality is equivalent to $0 < 1.01\epsilon$ and $\epsilon < 2.02\epsilon$, both of which are obvious. So we may assume $N>2$ and hence $(1+c)^{N-2} < (1.01)^{N-2}.$ So $(N-1)(1+c)^{N-2}\epsilon < (N-1)(1.01)^{N-2}\epsilon.$ So we just need to show $(N-1)(1.01)^{N-2}\epsilon < 1.01N\epsilon\Leftrightarrow (1.01)^{N-3} < \frac{N}{N-1}.$ Let $f(x) = \ln x - \ln (x-1) - (x-3)\ln 1.01$. Clearly $f(3) > 0$. For $x\ge 3$, we have $f'(x) = \frac{1}{x(x-1)} - \ln 1.01.$


Fact 1: Let $c, x$ be real numbers such that $0 < c < 1/100$ and $1 \le x \le \frac{1}{100c}$. Then
$$(1 + c)^{x - 1} - 1 < \frac{101}{100} x c.$$
(The proof is given at the end.)
According to Fact 1, the desired result follows.

Proof of Fact 1:
Denote $q = 101/100$.
Taking logarithm, it suffices to prove that
$$f(x) := \ln(1 + q x c) - (x - 1)\ln (1 + c)  > 0.$$
Let
$$t := \frac{1/(100c) - x}{1/(100c) - 1} \in [0, 1].$$
We have $x = t\cdot 1 + (1 - t)\cdot 1/(100c)$.
Since $f(x)$ is concave (easy to prove), we have
$$f(x) = f(t\cdot 1 + (1 - t)\cdot 1/(100c))
\ge t f(1) + (1 - t)f(1/(100c)).$$
Also, we have $f(1) = \ln(1 + qc) > 0$ and
\begin{align*}
 f(1/(100c)) &= \ln(1 + q/100) - (1/(100c) - 1)\ln(1 + c)\\
 &\ge \ln(1 + q/100) - (1/(100c) - 1)c \tag{1}\\
 &= \ln(1 + q/100) - 1/100 + c \\
 &> \ln(1 + q/100) - 1/100 \\
 &= \ln(1 + 101/10000) - 1/100\\
 &> 0
\end{align*}
where we have used $\ln(1 + u) \le u$ for all $u \ge 0$ in (1).
Thus, $f(x) > 0$ for all $1 \le x \le 1/(100c)$.
We are done.
