Prove that a series of integrals converges to zero exponentially

Question

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

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.

Answer 1

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} &

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.