Calculating the remainder of the series $1/n!$ (euler number)

Question

I am currently studying series and in my textbook and there's an example of calculating the remainder of the series which I don't understand completely

The series in question is: $$\sum_{i=0}^\infty \frac{1}{n!} = e,$$

so the remainder is \begin{align} \sum_{k=n+1}^\infty \frac{1}{k!} &= \frac{1}{(n+1)!}\left(1 + \frac{1}{n+2} + \frac{1}{(n+2)(n+3)} + ...\right) \\&< \frac{1}{(n+1)!}\left(1+\frac{1}{(n+1)}+\frac{1}{(n+1)^2} + ...\right) \\&= \frac{1}{(n+1)!}\frac{1}{1-\frac{1}{(n+1)}} \\&= \frac{1}{n!n} \end{align}

What I don't understand and I need clarification about is the inequality part (why is the first term less than the second) and why the term after the inequality equals what the textbook says it equals.

Thanks in advance!

Answer 1

For $k > 1$ we have

$$\frac{1}{n+k}<\frac{1}{n+1}.$$

Answer 2

"why the term after the inequality equals what the textbook says it equals."

-> Check what is the sum of a geometric series with a reason of $\frac{1}{n+1}$