prove $F(p)$ is non-negative for all $p$ Let $p>0$. Define,  $$ \displaystyle F(p) = 1- \,\dfrac{\dfrac{p^N}{N!}\,(N+1)\,\left(\sum_{n=0}^N\, \dfrac{p^n}{n!}\right)- \,\dfrac{p^{N+1}}{N!}\,\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}}{\left(\sum_{n=0}^N\, \dfrac{p^n}{n!}\right)^2}\,\,\,-(*)$$
I need to prove that $F(p)>0$ for all $p>0$. The result holds when $p<1$ but I'm having difficulty proving it for $p\geq 1$.
My work (Assume $N\geq 3$ since the result holds for $N=1,2$):
Taking the common denominator and further simplifying the numerator of $F(p)$,
$$\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\left(\sum_{n=0}^{N}\,\dfrac{p^n}{n!}-\dfrac{p^N\,(N+1)}{N!}\,+\dfrac{p^{N+1}}{N!}\right)\,+ \dfrac{p^N}{N!} \, \sum_{n=0}^{N}\,\dfrac{p^n}{n!}\,-\left(\dfrac{p^N}{N!}\right)^2\,(N+1) $$
$$= \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\left(\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!} -\dfrac{p^N}{(N-1)!}\,+\dfrac{p^{N+1}}{N!}\right)\,+ \dfrac{p^N}{N!} \, \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\,-\dfrac{p\,^{2N}}{N!(N-1)!}$$
$$= \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\left(\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}-\dfrac{p^N}{(N-1)!}\right)\,+ \dfrac{p^{N+1}}{N!}\,\sum_{n=0}^{N-2}\, \dfrac{p^n}{n!}+ \dfrac{p^N}{N!} \, \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\,\, (**)$$
I'm not sure how to proceed any further. Any help would be appreciated.

Let $x = \sum_{n=0}^N \frac{p^n}{n!}$. It suffices to prove that
$$x^2 - \frac{p^N}{N!}(N+1)x + \frac{p^{N+1}}{N!}\left(x - \frac{p^N}{N!}\right) > 0$$
which is quadratic in $x$.
It suffices to prove that
$$x > \frac{p^N}{2\cdot N!} (N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1})$$
or
$$\sum_{n=0}^N \frac{p^n}{n!}
> \frac{p^N}{2\cdot N!} (N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1})$$
or
$$\mathrm{e}^{-p}\sum_{n=0}^N \frac{p^n}{n!}
> \frac{\mathrm{e}^{-p} p^N}{2\cdot N!} (N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1}).$$
Let $f(p) = \mathrm{LHS} - \mathrm{RHS}$, for fixed $N$.
By using $(\mathrm{e}^{-p}\sum_{n=0}^N \frac{p^n}{n!})' = - \frac{1}{N!}p^N\mathrm{e}^{-p}$, we have
\begin{align}
f'(p) &= - \frac{1}{N!}p^N\mathrm{e}^{-p}\\
&\quad - \frac{\mathrm{e}^{-p} p^N(N-p)}{2p\cdot N!}(N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1})\\
&\quad - \frac{\mathrm{e}^{-p} p^N}{2\cdot N!}\left(-1 + \frac{p-N + 1}{\sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1}}\right)\\
&= - \frac{\mathrm{e}^{-p} p^N}{2pQ\cdot N!}\Big((N^2-2Np+p^2+N)Q + B \Big)
\end{align}
where $Q = \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1}$
and $B = N^3-3N^2p+3Np^2-p^3+2N^2-Np-p^2+N$.
Since $(N^2-2Np+p^2+N)^2Q^2 - B^2 = 4Np^2 > 0$,
we have $(N^2-2Np+p^2+N)Q + B > 0$ and thus $f'(p) < 0$.
Also, clearly, $\lim_{p\to \infty} f(p) = 0$. Thus, $f(p) > 0$. We are done.
