An = [2^(n-1) + 1] / [3^(n-1) + 1]
It converges to 0. Why is this?
It converges to 0. Why is this?
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take out the ones for starters as the limit goes to infinity and sum constants don't effect the function (much)
An lim->inf [2^(n-1)/(3^(n-1)]
now which one grows faster 2^(n-1) or 3^(n-1)? well seeming as how the coefficient is larger in 3^(n-1) it will grow at a faster rate. So the function will go to zero because the denominator grows at a faster rate than the numerator. This is true everytime this occurs.
An lim->inf [2^(n-1)/(3^(n-1)]
now which one grows faster 2^(n-1) or 3^(n-1)? well seeming as how the coefficient is larger in 3^(n-1) it will grow at a faster rate. So the function will go to zero because the denominator grows at a faster rate than the numerator. This is true everytime this occurs.
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Because the denominator grows faster than the numerator as "n" increases.