Cfrgtkky

This makes sense intuitively but I am not quite sure how to prove it rigorously. Any hints on how to proceed?

I assume $lim_na_n=0$ otherwise it is trivial: then
$$
lnPi_nbig({1over{1+ a_n}}big)=sum -ln(1+a_n),
$$
and this is equivalent to $-sum a_n$ since
$$ln(1+a_n)simeq a_ntext{ if }lim_na_n=0.$$

Let $a_ngeq 0$ for all $n.$ By induction on $n$ we have $$prod_{j=1}(1+a_j)geq 1+sum_{j=1}^na_j.$$ So if the sum diverges then so does the product.

We have $exp(a_j)geq 1+a_j, $ so $$exp (sum_{j=1}^na_j)geq prod_{j=1}^n(1+a_j).$$ So if the sum converges then so does the product.

Fact 1: For any $xgeq0$, we have $ln(1+x)leq x$.

Proof: Let $f:[0,infty)rightarrowmathbb{R}$ be defined by $f(x)=x-ln(1+x)$.
Observe that $f'(x)=frac{x}{1+x}geq0,$ so $f$ is increasing. For
any $xin[0,infty)$, we have $f(x)geq f(0)=0$.

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Fact 2: For any $xin[0,1]$, we have $ln(1+x)geqfrac{x}{2}$.

Proof: Let $g:[0,1]rightarrowmathbb{R}$ be defined by $g(x)=ln(1+x)-frac{x}{2}$.
Observe that $g'(x)=frac{1-x}{2(1+x)}geq0$, so $g$ is increasing.
For any $xin[0,1]$, we have $g(x)geq g(0)=0$.

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Let $P_{n}=prod_{k=1}^{n}frac{1}{1+a_{k}}$. Then
begin{eqnarray*}
-ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
& leq & sum_{k=1}^{n}a_{k}.
end{eqnarray*}
If $P_{n}rightarrow0$, we have $-ln P_{n}rightarrowinfty$, so
$sum_{k=1}^{infty}a_{k}=infty$.

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For each $k$, let $b_{k}=min(a_{k},1)leq a_{k}$. Then,
begin{eqnarray*}
-ln P_{n} & = & sum_{k=1}^{n}ln(1+a_{k})\
& geq & sum_{k=1}^{n}ln(1+b_{k})\
& geq & frac{1}{2}sum_{k=1}^{n}b_{k}.
end{eqnarray*}
If $sum_{k=1}^{infty}a_{k}=infty$, we will have $sum_{k=1}^{infty}b_{k}=infty$.
(For, consider two cases. Case 1: There exists $N$ such that $a_{k}leq 1$
whenever $kgeq N$. In this case, $sum_{k=1}^{infty}b_{k}geqsum_{k=N}^{infty}b_{k}=sum_{k=N}^{infty}a_{k}=infty$.
Case 2: There does not exist $N$ such that $a_{k}leq 1$ whenever $kgeq N$.
Then, there exists a subsequence $(a_{k_{l}})_{l}$ such that $a_{k_{l}}>1$.
We have that $sum_{k=1}^{infty}b_{k}geqsum_{l=1}^{infty}b_{k_{l}}=sum_{l=1}^{infty}1=infty$.)
Therefore, $-ln P_{n}rightarrowinfty$ and hence $exp(-ln P_{n})rightarrowinfty$.
But $exp(-ln P_{n})=frac{1}{P_{n}}$. It follows that $P_{n}rightarrow0$.