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I would like to estimate the scale of the following series,

$$S(m,t)=e^{-(m+1) t} \sum _{k=0}^{\infty } \frac{t^k}{k!}\left(\sum _{r=0}^k \frac{t^r}{r!}\right)^{m},$$

where $e$ is the base of natural logarithm, $m$ is an nonnegative integer and $t$ is a positive real number.

I turned the equation into another form,

$$S(m,t) = e^{-t} \sum _{k=0}^{\infty } \frac{t^k }{k!}\left(\frac{\Gamma (k+1,t)}{k!}\right)^m.$$ I noticed that $S(0,t)$ = 1, $S(1,t)$ is highly related to the existing question Cauchy product on exponential-looking power series, and $$S(m_2,t)<S(m_1,t)<1\,\text{ if }\,( 1<m_1<m_2).$$ But I cannot go further in giving an appropriate scale of $S(m,t)$.

Is it possible that $S(m, t)$ can be estimated as a polynomial of $t$? I am interested in giving an estimate of $S(m,t)$ as $S(m,t) = O (t^{-\alpha m}),(\alpha >0 )$ or something like that.
Would anyone give me any clue on this? Thanks in advance!

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up vote 2 down vote accepted

It turns out that $S(m,t)=P(A)$ where $A=[X_{m+1}\geqslant\max\{X_i;1\leqslant i\leqslant m\}]$ and $(X_i)_{1\leqslant i\leqslant m+1}$ is i.i.d. with Poisson distribution with parameter $t$. To see this, note that, for every $k\geqslant0$, $$ P(A\mid X_{m+1}=k)=P(X_1\leqslant k)^m=\left(\mathrm e^{-t}\sum_{r=0}^k\frac{t^r}{r!}\right)^m, $$ and that $$ P(A)=E(P(A\mid X_{m+1}))=\sum_{k=0}^{+\infty}\mathrm e^{-t}\frac{t^k}{k!}P(A\mid X_{m+1}=k). $$ Let $\mathbf I=\{1,2,\ldots,m+1\}$, $M=\max\{X_i;i\in\mathbf I\}$, $A_J=[\forall i\in J, X_i=M]$ for every $J\subseteq\mathbf I$, and $B_i=A_{\{1,2,\ldots,i\}}$ for every $i$ in $\mathbf I$. Then $P(A_J)=P(B_i)$ for every $J\subseteq\mathbf I$ of size $i$ and $$ S(m,t)=P(B_1). $$ The union of the sets $A_{\{i\}}$ for $i$ in $\mathbf I$ is the universe hence $(m+1)P(B_1)\geqslant1$, that is, $$ S(m,t)\geqslant\frac1{m+1}. $$ In the other direction, the exact inclusion-exclusion formula $$ 1=\sum_{i=1}^{m+1}(-1)^{i+1}{m+1\choose i}P(B_i), $$ yields the inequality $1\geqslant (m+1)P(B_1)-\frac12m(m+1)P(B_2)$ hence $$ S(m,t)\leqslant\frac1{m+1}+\frac12mP(B_2). $$ At this point, one can write down $P(B_2)$ as the sum of the obvious series but I fail to see how to estimate it accurately... Maybe you have an idea?

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Very impressive answer, but still I have no idea how to estimate this series…Therefore, I raised another question hoping you could help me out. What is the probability of the number 1 and number 2 employees getting the bonus at a call center? –  Erdos Yi Jun 19 '13 at 0:22
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