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Some heuristic way to the answer, related to well-known properties of the entropy of probability distributions, is loosely as follows.

Set $m=\sum_i a_i$ and $p_i=a_i/m$, so that $p=(p_i)_{i\ge1}$ is a probability distribution with expectation $\langle p\rangle=n/m\ge 1$.

Then $$\ln\frac{(\sum a_i) !}{\Pi a_i!}\simeq n\frac{\mathcal H(p)}{\langle p\rangle},$$ in which $\mathcal H(p)=-\sum_i p_i\ln(p_i)$ is the entropy of $p$. For a given expectation $\langle p\rangle=1/\lambda\ge 1$, the maximal entropy is that of the geometric distribution $\Lambda_i=(1-\lambda)^{i-1}\lambda$, namely $$\mathcal H(\Lambda)=-\ln\lambda+(1-\tfrac1\lambda)\ln(1-\lambda),$$ leading to $$\frac{\mathcal H(\Lambda)}{\langle \Lambda\rangle}=-(\lambda\ln\lambda+(1-\lambda)\ln(1-\lambda))\le\ln 2$$ with equality when $\lambda=1/2$, and $a_i/n=p_i m/n=(1-\lambda)^{i-1}\lambda^{2}=2^{-i-1}$.

Roadblocks are:

1. the accuracy of the approximation by $n\,\tfrac{\mathcal H(p)}{\langle p\rangle}$,

2. the fact that one can achieve $\forall i\ge 1, a_i/n\simeq 2^{-i-1}$ only approximately,

3. ...

Some heuristic way to the answer, related to well-known properties of the entropy of probability distributions, is loosely as follows.

Set $m=\sum_i a_i$ and $p_i=a_i/m$, so that $p=(p_i)_{i\ge1}$ is a probability distribution with expectation $\langle p\rangle=n/m\ge 1$.

Then $$\ln\frac{(\sum a_i) !}{\Pi a_i!}\simeq n\frac{\mathcal H(p)}{\langle p\rangle},$$ in which $\mathcal H(p)=-\sum_i p_i\ln(p_i)$ is the entropy of $p$. For a given expectation $\langle p\rangle=1/\lambda\ge 1$, the maximal entropy is that of the geometric distribution $\Lambda_i=(1-\lambda)^{i-1}\lambda$, namely $$\mathcal H(\Lambda)=-\ln\lambda+(1-\tfrac1\lambda)\ln(1-\lambda),$$ leading to $$\frac{\mathcal H(\Lambda)}{\langle \Lambda\rangle}=-(\lambda\ln\lambda+(1-\lambda)\ln(1-\lambda))\le\ln 2$$ with equality when $\lambda=1/2$, and when $a_i/n=p_i m/n=(1-\lambda)^{i-1}\lambda^{2}=2^{-i-1}$.

Roadblocks are:

1. the accuracy of the approximation by $n\,\tfrac{\mathcal H(p)}{\langle p\rangle}$,

2. the fact that one can achieve $\forall i\ge 1, a_i/n\simeq 2^{-i-1}$ only approximately,

3. ...

Some heuristic way to the answer, related to well-known properties of the entropy of probability distributions, is loosely as follows :

set $m=\sum_i a_i$ and $p_i=a_i/m$, so that $p=(p_i)_{i\ge1}$ is a probability distribution with expectation $\langle p\rangle=n/m\ge 1$.

Then $$\ln\frac{(\sum a_i) !}{\Pi a_i!}\simeq n\frac{\mathcal H(p)}{\langle p\rangle},$$ in which $\mathcal H(p)=-\sum_i p_i\ln(p_i)$ is the entropy of $p$. For a given expectation $\langle p\rangle=1/\lambda\ge 1$, the maximal entropy is that of the geometric distribution $\Lambda_i=(1-\lambda)^{i-1}\lambda$, namely $$\mathcal H(\Lambda)=-\ln\lambda+(1-\tfrac1\lambda)\ln(1-\lambda),$$ leading to $$\frac{\mathcal H(\Lambda)}{\langle \Lambda\rangle}=-(\lambda\ln\lambda+(1-\lambda)\ln(1-\lambda))\le\ln 2$$ with equality when $\lambda=1/2$, and $a_i/n=p_i m/n=(1-\lambda)^{i-1}\lambda^{2}=2^{-i-1}$.

Some heuristic way to the answer, related to well-known properties of the entropy of probability distributions, is loosely as follows.

Set $m=\sum_i a_i$ and $p_i=a_i/m$, so that $p=(p_i)_{i\ge1}$ is a probability distribution with expectation $\langle p\rangle=n/m\ge 1$.

Then $$\ln\frac{(\sum a_i) !}{\Pi a_i!}\simeq n\frac{\mathcal H(p)}{\langle p\rangle},$$ in which $\mathcal H(p)=-\sum_i p_i\ln(p_i)$ is the entropy of $p$. For a given expectation $\langle p\rangle=1/\lambda\ge 1$, the maximal entropy is that of the geometric distribution $\Lambda_i=(1-\lambda)^{i-1}\lambda$, namely $$\mathcal H(\Lambda)=-\ln\lambda+(1-\tfrac1\lambda)\ln(1-\lambda),$$ leading to $$\frac{\mathcal H(\Lambda)}{\langle \Lambda\rangle}=-(\lambda\ln\lambda+(1-\lambda)\ln(1-\lambda))\le\ln 2$$ with equality when $\lambda=1/2$, and $a_i/n=p_i m/n=(1-\lambda)^{i-1}\lambda^{2}=2^{-i-1}$.

Roadblocks are:

1. the accuracy of the approximation by $n\,\tfrac{\mathcal H(p)}{\langle p\rangle}$,

2. the fact that one can achieve $\forall i\ge 1, a_i/n\simeq 2^{-i-1}$ only approximately,

3. ...

Some heuristic way to the answer, connected to a classic problem about the entropy, is loosely as follows : set $m=\sum_i a_i$ and $p_i=a_i/m$, so that $p=(p_i)_{i\ge1}$ is a probability distribution with expectation $\langle p\rangle=n/m\ge 1$. Then $$\ln\frac{(\sum a_i) !}{\Pi a_i!}\simeq n\frac{\mathcal H(p)}{\langle p\rangle},$$ in which $\mathcal H(p)=-\sum_i p_i\ln(p_i)$ is the entropy of $p$. For a given expectation $\langle p\rangle=1/\lambda\ge 1$, the maximal entropy is that of the geometric distribution $\Lambda_i=(1-\lambda)^{i-1}\lambda$, namely $$\mathcal H(\Lambda)=-\ln\lambda+(1-\tfrac1\lambda)\ln(1-\lambda),$$ leading to $$\frac{\mathcal H(\Lambda)}{\langle \Lambda\rangle}=-(\lambda\ln\lambda+(1-\lambda)\ln(1-\lambda))\le\ln 2$$ with equality when $\lambda=1/2$, and $a_i/n=p_i m/n=(1-\lambda)^{i-1}\lambda^{2}=2^{-i-1}$.

Some heuristic way to the answer, related to well-known properties of the entropy of probability distributions, is loosely as follows : 

set $m=\sum_i a_i$ and $p_i=a_i/m$, so that $p=(p_i)_{i\ge1}$ is a probability distribution with expectation $\langle p\rangle=n/m\ge 1$. 

Then $$\ln\frac{(\sum a_i) !}{\Pi a_i!}\simeq n\frac{\mathcal H(p)}{\langle p\rangle},$$ in which $\mathcal H(p)=-\sum_i p_i\ln(p_i)$ is the entropy of $p$. For a given expectation $\langle p\rangle=1/\lambda\ge 1$, the maximal entropy is that of the geometric distribution $\Lambda_i=(1-\lambda)^{i-1}\lambda$, namely $$\mathcal H(\Lambda)=-\ln\lambda+(1-\tfrac1\lambda)\ln(1-\lambda),$$ leading to $$\frac{\mathcal H(\Lambda)}{\langle \Lambda\rangle}=-(\lambda\ln\lambda+(1-\lambda)\ln(1-\lambda))\le\ln 2$$ with equality when $\lambda=1/2$, and $a_i/n=p_i m/n=(1-\lambda)^{i-1}\lambda^{2}=2^{-i-1}$.