Prove that a series of integrals converges to zero exponentially I have the following series, which I want to show converges to $0$ exponentially fast in $N$:
$$S_N=\sum_{n = N}^{\infty}\frac{1}{n!}M^{n}\int_{0}^{1}e^{- Mv}v^{n}\left(1-v^{r}\right)^{n}\frac{dv}{\sqrt{v}}\,,$$
with any given $r\ge1$ an integer, and $M$ is a constrained parameter that may be chosen appropriately to prove our statement. So, we need to show that there exists a choice $M\left( N,r  \right)$ for which 
$$S_N =\mathcal{O}\left( C^N\right) \,,\quad \left|C\right|<1 \,.$$
The constraint on $M$ is that I want the theorem to hold for sufficiently large $M$, i.e. that the proof will only require a minimal value of $M$. Specifically, if this can be shown for $M(N,r)\ge\left( \log r+ r \log 2\right) N$, then this is sufficient for me.
Indeed, for $r=1$ this is relatively straight-forward:
$$
\begin{align*}
S_N & = \sum_{n=N}^{\infty}\frac{M^{n}}{n!}\int_{0}^{1}e^{-Mv}v^{n}\left(1-v\right)^{n}\frac{dv}{\sqrt{v}}\\
 & <\sum_{n=N}^{\infty}\frac{M^{n}}{n!}\int_{0}^{1}e^{-Mv}v^{n}e^{-nv}\frac{dv}{\sqrt{v}}\\
 & =\sqrt{\frac{1}{M}}\sum_{n=N}^{\infty}\left[\frac{M}{M+n}\right]^{n+\frac{1}{2}}\int_{0}^{M+n}e^{-u}u^{n-\frac{1}{2}}\frac{du}{n!}\\
 & <\sqrt{\frac{1}{M}}\sum_{n=N}^{\infty}\left[\frac{M}{M+n}\right]^{n+\frac{1}{2}}\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!}\\
 & <\sqrt{\frac{1}{M}}\sum_{n=N}^{\infty}\left[\frac{M}{M+n}\right]^{n+\frac{1}{2}}\frac{1}{\sqrt{n}}\\
 & <\sqrt{\frac{1}{M}}\frac{1}{\sqrt{N}}\sum_{n=N}^{\infty}\left[\frac{M}{M+N}\right]^{n+\frac{1}{2}}\\
 & =\sqrt{\frac{1}{M}}\sqrt{\frac{M}{N\left(M+N\right)}}\frac{\left[\frac{M}{M+N}\right]^{N}}{1-\left[\frac{M}{M+N}\right]}\\
 & <\sqrt{\frac{M+N}{N^{3}}}\left[\frac{M}{M+N}\right]^{N}\,.
\end{align*}
$$
so we have $C=\frac{M}{M+N}<1$ for any choice of $M\left(N\right)$, as required.
However, I don't see a similar approach to show this claim for $r>1$. For $r=2$ I thought about again bounding $(1-v^2)^n<e^{-nv^2}$, so now  the integral may be expressed as some form of Mill's ratio, but I couldn't find any inequalities for it which include the additional factor of $v^n$ in the integrand.
The difficult point is to somehow bound the integrals but not too loosely, so that the sum of the bound over $n$ is convergent, and may be itself bounded by some behavior in $N$.
It is worth mentioning that I checked this claim numerically, up to r = 4, and it seems to hold.
 A: This can be shown by bounding the integrand:
$$
\begin{align*}
S_N & = \sum_{n=N}^{\infty}\frac{M^{n}}{n!}\int_{0}^{1}e^{-Mv}v^{n}\left(1-v^r\right)^{n}\frac{dv}{\sqrt{v}} \\
& \le \sum_{n=N}^{\infty}\frac{M^{n}}{n!}\int_{0}^{1}e^{-Mv-nv^r}v^{n-\frac{1}{2}}dv 
\end{align*}
$$
Noting that $v^{r}$ is a concave function, it is bounded by any line
drawn tangential to it. Denoting by $v_{0}>0$ the arbitrary point
where we draw the tangent, we can bound
\begin{align*}
v^{r} & <rv_{0}^{r-1}\left(v-v_{0}\right)+v_{0}^{r}=rv_{0}^{r-1}v-\left(r-1\right)v_{0}^{r}\,.
\end{align*}
Plugging this in, we have
\begin{align*}
S_N & <\sum_{n=N}^{\infty}\frac{1}{n!}M^{n}\int_{0}^{1}e^{-Mv-nrv_{0}^{r-1}v+n\left(r-1\right)v_{0}^{r}}v^{n-\frac{1}{2}}dv\\
 & <\sum_{n=N}^{\infty}\frac{1}{n!}M^{n}e^{n\left(r-1\right)v_{0}^{r}}\int_{0}^{\infty}e^{-\left(M+nrv_{0}^{r-1}\right)v}v^{n-\frac{1}{2}}dv\\
 & =\sum_{n=N}^{\infty}\frac{1}{n!}M^{n}e^{n\left(r-1\right)v_{0}^{r}}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\left(M+nrv_{0}^{r-1}\right)^{n+\frac{1}{2}}}\\
 & <\sqrt{\frac{1}{\left(M+Nrv_{0}^{r-1}\right)}}\sum_{n=N}^{\infty}\frac{1}{\sqrt{n}}\left(\frac{Me^{\left(r-1\right)v_{0}^{r}}}{M+nrv_{0}^{r-1}}\right)^{n}\\
 & <\sqrt{\frac{1}{N\left(\alpha+rv_{0}^{r-1}\right)}}\sum_{n=N}^{\infty}\left(\frac{\alpha e^{\left(r-1\right)v_{0}^{r}}}{\alpha+rv_{0}^{r-1}}\right)^{n}\,,
\end{align*}
where he have denoted $M=\alpha N$. This is a geometric series which
only converges if its quotient is smaller than one, namely if 
\begin{equation}
\alpha\left(e^{\left(r-1\right)v_{0}^{r}}-1\right)-rv_{0}^{r-1}<0\,.
\end{equation}
This condition is trivially satisfied for $r=1$. For $r>1$,
we note that for $v_{0}=0$ the left-hand side of this condition
equals $0$. However, since $e^{\left(r-1\right)v_{0}^{r}}$ can be
expanded in powers of $v_{0}^{r}$, the first non-vanishing derivative
of the lhs is the $\left(r-1\right)^{\text{th}}$, giving $\left(-r!\right)<0$.
Thus, this equation is satisfied at least in the neighborhood of $0^{+}$.
Indeed, for $v_{0}\ll1$, we can expand 
$$
\alpha\left(r-1\right)v_{0}^{r}-rv_{0}^{r-1}<0\,,
$$
giving us a self-consistent solution as long as $v_{0}<\frac{r}{\left(r-1\right)\alpha}$. 
To summarize, this implies that for any value of $M$, there exists a choice for $v_0$ for which the quotient of the geometric series is smaller than unity, i.e. some $C<1$, so that the sum converges and behaves is $C^N$ which tends to zero exponentially.
