Find analytic solution: for what $n_0$, $\forall n > n_0$ : $n \cdot p ^{n-1} \leq \delta$ holds?

Question

Suppose $p$ is a fixed number in $(0, 1)$ and $\delta$ is a small positive number s.t. $ 0 < \delta < p$.

What is $n_0$ such that for any $n > n_0$, the following holds: $$ n \cdot p ^{n-1} \leq \delta $$

Answer 1

Set $p:= e^{-y}$, where $0 <y$, real

Now consider:

$n e^{-y(n-1)}=\dfrac{n}{e^{y(n-1)}} =$

$\dfrac{n}{1+ y(n-1)+ y^2(n-1)^2/2 +....}$

$ \lt \dfrac{2n}{y^2(n-1)^2} \lt$

$(1/y^2)\dfrac{2n}{(n/2)^2}= (8/y^2)(1/n).$

Let $\epsilon >0$ be given.

Archimedean principle:

There is a $n_0$, positive integer with

$n_0 > (1/\epsilon)(8/y^2).$

For $n\ge n_0$ we have

$ne^{-y(n-1)} \lt (8/y^2)(1/n)\le (8/y^2)(1/n_0) \lt \epsilon.$

Used: $(n-1)^2 >(n/2)^2$ , $n \ge 3.$

Answer 2

$p,q\in(0,1)$, $p=1-q$, (Pick $q\in(0,1)$ so that this equality holds). We get $\delta \geq p^{n-1}n=(1-q)^{n-1}n\geq(1-(n-1)q)n $ (using Bernouli`s inequality). $\delta \geq n-n^2q-qn \leftrightarrow n^2q+(q-1)n+\delta\geq0$. From here you can find the roots of the polynomial (Relative to $n$), take the bigger one ($n_0$), and since this is a parabola, the inequality holds for all $n\geq n_0$

Answer 3

Let $r=\frac {1}{p}-1.$ That is, $p=\frac {1}{1+r}.$ We have $r>0.$

If $n\geq 4$ then by the Binomial Theorem we have $$\frac {1}{p^{n-1}}=(1+r)^{n-1}=1+r(n-1)+r^2\frac {(n-1)(n-2)}{2}+...>$$ $$>r^2\frac {(n-1)(n-2)}{2}$$ so we have $$\frac {1}{np^{n-1}}>r^2 \frac {(n-1)(n-2)}{2n}=\frac {r^2}{2}(n-3+2/n)>\frac {r^2}{ 2}( n-3)$$ so we have $np^{n-1}< \frac {2}{r^2(n-3)} .$ And the rest is obvious.

I felt like giving an answer using only the most elementary means (not even Bernoulli's Inequality).