This is a problem which I found in a math textbook some time ago. I tried solving it but I did not manage to solve it as quickly as I wanted and so I lost patience. Therefore I posted it in MathSE. I liked quite a lot one of the solutions I got, so I wrote it on a sheet of paper and kept it.
Here is the full story...
Problem:
Find limx→∞n√(x+a1)(x+a2)...(x+an)−x,
where x is a real variable and n is a fixed natural number n≥1.
Solution:
Let
f(x)=n√(x+a1)(x+a2)⋅(x+an)
We now calculate:
limx→∞f(x)x=limx→∞n√(1+a1x)(1+a2x)⋅(1+anx)=n√1=1
We now form the below difference and we rework it a bit:
f(x)−x=f(x)n−xn∑n−1k=0f(x)n−1−kxk=f(x)n−xnf(x)n−1+f(x)n−2x+⋯+f(x)xn−2+xn−1
We note here that the numerator P(x)=f(x)n−xn is a polynomial of degree n−1 and it has a leading coefficient of c=a1+a2+⋯+an. The denominator is composed of the sum of n terms of the form f(x)n−1−kxk (for k=0,1,…,n−1).
So we divide the numerator and the denominator by xn−1 and we get:
limx→∞f(x)−x=limx→∞f(x)n−xnxn−1f(x)n−1xn−1+f(x)n−2xn−2+⋯+f(x)x+1=limx→∞P(x)xn−1f(x)n−1xn−1+f(x)n−2xn−2+⋯+f(x)x+1=c1+1+⋯+1⏟n times=a1+a2+⋯+ann
The cool thing here is that the limit value happens to be the arithmetic mean of the numbers a1,a2,…an.
See also Math SE question/solution here
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