Cfrgtkky

Can someone suggest how to obtain an expression for $S[i]$ given that S[0] = 0,

$S[i]=S[i-1]*(1-gamma_i)^2 + gamma_i^2$ where $gamma_i = frac{g+1}{g+i}$

EDIT: $g>0$ and $g$ can be assumed to be a natural number.

An exact expression or a lower bound would be helpful.

Thanks!

I was able to determine a formula for this based off of looking at the first few elements of the sequence.

I have that $$s_i=gamma_i^2(sum_{n=1}^{i-1}(prod _{k=n}^{i-1}frac{k^2}{(g+k)^2})+1).$$

Now let’s look at bounding it from below. From your hunch that there is a $c$ such that $(g+i)s_i>c$, we proceed by finding which $c$ might give us an induction argument.

Suppose $(g+i)s_i>c$ for $i>0$. Then $$(g+i+1)s_{i+1}=frac{i^2 s_i}{(g+i+1)}+frac{(g+1)^2}{g+i+1} > frac{i^2 c+(g+1)^2}{(g+i+1)^2}.$$

If we find a $c$ such that the final expression of the previous paragraph is greater than $c$ for all $i>0$, then we have the desired result. After some algebra and working backwards, it can be shown that any $c<(g+1)/2$ satisfies it. Following through with the induction argument, we can see that the base case ($i=1$) is satisfied since $s_1>1/2$.

For a sequence $(S_i)_{iinmathbb N}$ generated as $S_i=a_iS_{i-1}+b_i$ with $a_i,b_iinmathbb R$, by recursively expanding once ore twice you can easily obtain a general formula.
$$begin{align}
S_i
{}={} &
a_iS_{i-1}+b_i
\
{}={} &
a_i(a_{i-1}S_{i-2}+b_{i-1})+b_i
\
& {}vdots{}
\
{}={} &
a_ia_{i-1}cdots a_1 S_0
{}+{}
b_i+a_ib_{i-1}+a_ia_{i-1}b_{i-2}+ldots+ a_ia_{i-2}cdots a_2b_1
\
{}={} &
S_0prod_{j=1}^ia_j
{}+{}
sum_{j=1}^ib_jprod_{k=j+1}^ia_k.
end{align}$$
In your case $S_0=0$, hence the first term above vanishes.
What you need is a way to compute the remaining products and sums.

Since
$
a_k
{}={}
left(1-frac{g+1}{g+k}right)^2
{}={}
frac{(k-1)^2}{(g+k)^2}
$
and
$b_j=frac{(g+1)^2}{(g+j)^2}$, you have
$$begin{align}
color{red}{b_j}prod_{k=j+1}^ia_k
{}={} &
color{red}{frac{(g+1)^2}{(g+j)^2}}
,,
frac{j^2}{(g+j+1)^2}
frac{(j+1)^2}{(g+j+2)^2}
{}cdots{}
frac{(i-2)^2}{(g+i-1)^2}
frac{(i-1)^2}{(g+i)^2}
\
{}={} &
color{red}{frac{(g+1)^2}{(g+j)^2}}
,,
frac{(i-1)!^2}{(j-1)!^2}
frac{(g+j)!^2}{(g+i)!^2}
\
{}={} &
(g+1)^2
frac{(i-1)!^2}{(j-1)!^2}
frac{(g+j-1)!^2}{(g+i)!^2},
end{align}$$
where for simplicity $g$ was assumed natural.
The general case can easily be adapted by redefining the factorial accordingly (that is, $g!:=g$ and $(x+1)!:=(x+1)x!$, defined for $xin g+mathbb N$).

Therefore,
$$begin{align}
S_i
{}={} &
frac{(g+1)^2(i-1)!^2}{(g+i)!^2}
sum_{j=0}^{i-1}
frac{(g+j)!^2}{j!^2}
\
tag{1}
mbox{(EDIT)}quad
{}={} &
frac{1}{binom{g+i}{i-1}^2}
sum_{j=0}^{i-1}
binom{g+j}{j}^2.
end{align}$$
Now, in order to obtain a lower bound, notice that
$$
binom{g+j}{j}
{}={}
binom{g+j-1}{j-1}frac{g+j}{j}
{}geq{}
binom{g+j-1}{j-1}left(textstyle 1+frac{g}{i-1}right)
quad
mbox{for }jleq i-1
$$
therefore,
$$
binom{g+j}{j}^2
{}geq{}
binom{g}{0}^2left(textstyle 1+frac{g}{i-1}right)^{2j}
{}={}
left(textstyle 1+frac{g}{i-1}right)^{2j},
$$
and plugging this into (1) yields
$$begin{align}
S_i
{}geq{} &
frac{1}{binom{g+i}{i-1}^2}
sum_{j=0}^{i-1}
left(textstyle 1+frac{g}{i-1}right)^{2j}
{}={}
frac{1}{binom{g+i}{i-1}^2}
frac{(1+frac{g}{i-1})^{2i}-1}{(1+frac{g}{i-1})^2-1}.
end{align}$$