Calculate $\lim_{a \to 0}\frac{2^n-(2-a)(2-2^2a)(2-3^2a).....(2-n^2a)}{2^na}$

Question

I have tried every way I know to calculate this end (divide $a^n$, divide $2^n$, derivation) but without any result.

Can you help me calculate it? and thank you very much.

$$ \lim_{a \to 0}\frac{2^n-(2-a)(2-2^2a)(2-3^2a).....(2-n^2a)}{2^na} $$

I could not find a solution

Answer 1

For fixed $n$, the expression in the numerator is a polynomial $f(a)=c_0+c_1a+c_2a^2+\ldots$ in $a$ (apparently of degree $\deg f=n$, so we stop at $c_na^n$). By plugging in $a=0$, we immediate see that $c_0=f(0)=2^n-(2-0)\cdots(2-0)=0$. Then it readily follows that $$ \lim_{a\to0}\frac{f(a)}{2^na}=\lim_{a\to0}\frac{c_1+c_2a+\ldots+c_na^{n-1}}{2^n}=\frac{c_1}{2^n}.$$ Thus we need the linear term of $f(a)$, or equivalently, the negative linear term of the product $(2-a)(2-2^2a)\cdots(2-n^2a)$. Note that for two polynomials $g(X)=g_0+g_1X+\ldots$ and $h(X)=h_0+h_1X+\ldots$, we have $$\tag1g(X)h(X)=g_0h_0+(g_0h_1+g_1h_0)X+\ldots$$ and this allows us to compute the linear and constant term of a product from only the linear and constant temrs of the factors. Letting $g_n(X)=(1-\frac12X)(1-\frac {2^2}2X)\cdots(1-\frac{n^2}2X)$, we have the recursion $g_n(X)=g_{n-1}(X)\cdot(1-\frac{n^2}2X)$. As clearly the constant term of $g_n$ is $1$, we find from $(1)$ the following recursion for the linear coefficient $g_{n,1}$ of $g_n(X)$: $$ g_{n,1}=g_{n-1,1}-\frac{n^2}2$$ with the initial condition $g_{0,1}=0$. Quite obviously, thius makes $$g_{n,1}=\frac12\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{12}.$$ Now as $f(X)=2^n-2^ng(X)$, we conclude $$\lim_{a\to0}\frac{f(a)}{2^na}=\frac{c_1}{2^n}=-g_{n,1}=\frac{n(n+1)(2n+1)}{12}. $$