Can you explain this expression in the book (a text-book of convergence "w.L FERRAR" )
The writer present this ideas
( if y is a fixed number graeter than unity, and k is any fixed positive integer, then the sequence $\frac{n^k}{y^n} $ converges to zero
Before we give the proof of this theorem, we try to explain some of the ideas that lead to the proof
i)...... if  l is a fixed number and n is to be thought of as a very large number, then n-l is about as big as n, further, extending this ideas a little, if k is a fixed number and n is to be thought of as a very large number, then
$n(n-1)(n-2).....(n-k)$
is about as big as $n^{k+1}$
(ii) the y of our Problem exceeds unity and we can write
$y^n=(1+p)^n$ p>0
The right-hand side can be expended, when n is a positive integer, by the binomial theorem, and, since each term of the expansion is positive, so if we take $(n>k+1)$
$y^n=(1+p)^n>\frac{n(n-1)....(n-k)}{(k+1)!}p^{k+1}$)
I can't understand  the last expression، so can you explain it?
And also I need more clarification for the partie (i)
 A: The last expression comes from the binomial expansion of $(1+p)^n$.
Remember that $$(1+p)^n=\sum^{n}_0\binom{n}i p^i$$
Now since each term on the RHS is positive, the sum is definitely greater than an individual term.
$$\sum^{n}_0\binom{n}i p^i\ge\binom{n}{k+1} p^{k+1}$$ which is what expands into the last expression.
As for the explanation of part (i) it can be understood as
$$\frac{\sum_{r=0}^{k} (n-r)}{n^{k+1}}=\sum_{r=0}^{k} (1-r/n)\rightarrow 1\ as\ n\rightarrow\infty$$. which justifies the approximation.
A: FYI - when people ask for clarification, just edit the post.  No need to keep deleting and redoing it.
To answer your question, if we have $y^n = (1+p)^n$ we can use the binomial theorem, which says that
$$(1+x)^{n} = \sum_{k=0}^{n}\binom{n}{k}x^{k}.$$  In this case, we have
$$y^n = (1+p)^n = \sum_{k=0}^{n}\binom{n}{k}p^{k}.$$  Now, so long as $n > k+1$, and since every term in the sum is positive, we can take just the term with the $k+1$ power and it will be less than the entire sum.  So, we get
\begin{align*}
y^n &= (1+p)^n\\[5pt]
&= \sum_{k=0}^{n}\binom{n}{k}p^{k}\\[5pt]
&=\binom{n}{1}p + \binom{n}{2} + \dotsb + \binom{n}{k+1}p^{k+1} + \dotsb + \binom{n}{n}p^{n}\\[5pt]
&> \binom{n}{k+1}p^{k+1}\\[5pt]
&= \frac{n!}{(k+1)!(n-k-1)!}p^{k+1}\\[5pt]
&= \frac{n(n-1)\dotsb(n-k)}{(k+1)!}p^{k+1}.
\end{align*}
As for part (i), if we think of $n$ as being very large, then $n-1 \approx n$, which means that $n(n-1) \approx n^2$.  Extending this idea, if $k$ is some fixed number, and $n$ is much larger than $k$, then the numbers $$n, n-1, n-2, \dotsc, n-k$$ are all approximately equal to $n$.  So, this gives us $$n(n-1)(n-2)\dotsb(n-k) \approx n^{k+1},$$ because there are $k+1$ terms in the product.
