Understanding why a sum is greater than another I'm currently going through a research paper and was trying to redo a proof on my side.
The paper states :
$$
\begin{align}
\text{P}[failure] &\le \text{P}[s + 1 \text{ runs of A fail}] \\
&\le \sum_{i \ge s + 1} \binom{2s + 1}{i} \left(\frac{1}{4}\right)^i \left(\frac{3}{4}\right)^{2s + 1 - i} \\
&\le \left(\frac{1}{4}\right)^{s + 1} \left(\frac{3}{4}\right)^s \sum_{i \ge s + 1} \binom{2s + 1}{i}
\end{align}
$$
I just can't seem to understand why one can move from the second to the last line. I understand that we use the frontier $i = s + 1$ and replace, but I don't get why this means we can extract the fractions from the sum, and why the expression is larger than the line before.
What is the obvious thing I'm missing ?

*

*You probably have a typo in the exponent of $ \frac{3}{4} $, it should be $ s $.

*What you are asking is equivalent to why is
$$ \left( \frac{1}{4} \right)^a \ge \left( \frac{1}{4} \right)^b, $$
given that $ 0 < a \le b $.

*Try to understand the inequality with everything inside the sums, that is
\begin{equation}
\sum_{i \ge s + 1} \binom{2s + 1}{i} \left(\frac{1}{4}\right)^i \left(\frac{3}{4}\right)^{2s + 1 - i} \le \sum_{i \ge s + 1} \binom{2s + 1}{i}\left(\frac{1}{4}\right)^{s + 1} \left(\frac{3}{4}\right)^s 
\end{equation}
Since the last two terms do not depend on $ i $, we can simply factor them out of the sum - same way we can write
$ a x + b x + c x = x(a+b+c)$
