Polynomial Limit

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#1 h Apr 23, 2025, 11:55 AM

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If $P_{n}(x) = \prod_{k=1}^{n} \left(x + \frac{1}{2^k}\right) = \sum_{k=0}^{n} a_{k} x^k$ then find the value of $\lim_{n \to \infty} \frac{a_{n - 2}}{a_{n - 4}}.$

This post has been edited 2 times. Last edited by P162008, Apr 25, 2025, 1:25 AM
Reason: Typo

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#2 h Apr 24, 2025, 10:25 PM • 1 Y

Y by P162008

I assume it should be $\Pi_{k=1}^n$ since then the degree is $n$ as it should be.
We have $\Pi_1^n (x+1/2^k)=\Pi_1^n (x-t_k)=\sum_0^n a_k x^k$ and we will determine the limits of $a_{n-2},a_{n-4}$ separately.
We have $a_{n-2}=\sum_{1\le k<l\le n} t_kt_l=\sum_{1\le k<l\le n} {1\over {2^{k+l}}}\rightarrow \sum_{k<l}{1\over {2^{k+l}}}=\sum_{k=1}^\infty (1/2)^k\sum_{l=k+1}^\infty (1/2)^l=$
$\sum_1^\infty (1/4)^k={{1/4}\over {1-1/4}}=1/3$ .
In the same way $a_{n-4}\rightarrow \sum_{p<q<r<s} (1/2)^{p+q+r+s}=\sum_{p=1}^\infty (1/2)^p \sum_{q=p+1}^\infty (1/2)^q \sum_{r=q+1}^\infty (1/2)^r \sum_{s=r+1}^\infty (1/2)^s=1/315$ .
I leave the details of the last part to the reader, just work your way from inside to outside.
So we get 105.

This post has been edited 1 time. Last edited by alexheinis, Apr 24, 2025, 10:27 PM

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