Consider the sequence: (a,sub(n))={1/n E(k=1 to n) 1/1+(k/n)}

Show that the limit(as n-> infinity) A(sub(n))= ln 2 by interpreting a(sub(n)) as a Reimann Sum of a Definite Integral.

To show that the limit of A(sub(n)) as n approaches infinity is ln 2, we need to interpret the sequence (a,sub(n)) as a Riemann sum of a definite integral.

First, let's rewrite the sequence (a,sub(n)) using the given formula:

(a,sub(n)) = (1/n) * ∑ (k=1 to n) 1 / (1 + k/n)

By observing the pattern, we can notice that each term of the summation is in the form of 1/(1 + x), where x is a constant. Let's manipulate the expression to better represent that:

(a,sub(n)) = (1/n) * ∑ (k=1 to n) 1 / (1/n + k/n)

Now, we can rewrite this expression in terms of x by substituting x = k/n:

(a,sub(n)) = (1/n) * ∑ (x=1/n to 1) 1 / (1/n + x/n)

Next, let's rewrite the summation as a Riemann sum. We can do this by considering a partition of the interval [1/n, 1], dividing it into n subintervals of equal width Δx = 1/n. We denote the left endpoint of each subinterval as x(sub(i)) and choose a sample point c(sub(i)) within each subinterval.

With the partition and sample points defined, we can rewrite the expression as a Riemann sum:

(a,sub(n)) = ∑ (i=1 to n) 1 / (1/n + c(sub(i))/n) * Δx

Now, if we let n approach infinity, the width of each subinterval Δx approaches 0, and the Riemann sum becomes a definite integral:

lim(n->∞) (a,sub(n)) = ∫ (1/(1 + x)) dx (from x = 0 to x = 1)

To evaluate this integral, we can integrate the function 1/(1 + x) with respect to x:

lim(n->∞) (a,sub(n)) = [ln(1 + x)] evaluated from x = 0 to x = 1
lim(n->∞) (a,sub(n)) = ln(1 + 1) - ln(1 + 0)
lim(n->∞) (a,sub(n)) = ln(2) - ln(1)
lim(n->∞) (a,sub(n)) = ln(2)

Therefore, the limit of (a,sub(n)) as n approaches infinity is ln(2), which verifies that A(sub(n)) approaches ln(2) as n approaches infinity.