Power series, part of proof explanation.

Question

I am struggling to understand the red inequality $\color{red}{\leq}$.

how one could just change $k$ to $2$, and then change the sum previously going from $2$ to n, to a sum going from $0$ to $n$. Now it might very well be that there are some summation rules that I have forgotten about, thanks for the help.

Therefore, if $|h|\leq H$ we have \begin{align*}|(x+h)^n-x^n-nx^{n-1}h| =& \left|\sum_{k=2}^n\binom{n}{k}x^{n-k}h^k\right|\\ \leq &\sum_{k=2}^n\binom{n}{k}|x|^{n-k}\frac{|h|^k}{H^k}H^k \\ \color{red}{\leq}&\frac{|h|^2}{H^2}\sum_{k=0}^n\binom{n}{k}|x|^{n-k}H^k\\=&\frac{|h|^2}{H^2}\big(|x|+H\big)^n\end{align*}

Answer 1

Since $|h|\leq H$ we have $\frac{|h|}{H}\leq 1$ and thus $$\frac{|h|^k}{H^k}\leq \frac{|h|^2}{H^2}$$ for every $k\geq 2$. Moreover adding two nonnegative terms (since $0\leq H$) in the sum will also make it larger so that

$$\sum_{k=2}^n\binom{n}{k}|x|^{n-k}\frac{|h|^k}{H^k}H^k \leq \sum_{k=2}^n\binom{n}{k}|x|^{n-k}\frac{|h|^2}{H^2}H^k=\frac{|h|^2}{H^2}\sum_{k=2}^n\binom{n}{k}|x|^{n-k}H^k\\ \leq \frac{|h|^2}{H^2}\left(\sum_{k=2}^n\binom{n}{k}|x|^{n-k}H^k +\binom{n}{1}|x|^{n-1}H^1+\binom{n}{0}|x|^{n-0}H^0\right) = \frac{|h|^2}{H^2}\sum_{k=0}^n\binom{n}{k}|x|^{n-k}H^k$$

Answer 2

Observe that each summand has a factor of $\left(\dfrac{|h|}{H}\right)^2$. First take that factor outside. Then as $\dfrac{|h|}{H}\leq1$, erasing its powers that appear as factors in the sum will increase the value of the expression. Finally adding two more terms will increase the value once more. Observe that all these are done in order to get a clean upper bound that is easy to manipulate.