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 can interpret A(sub(n)) as a Riemann sum of a definite integral.
First, let's rewrite the expression for A(sub(n)):
A(sub(n)) = 1/n * Σ (k=1 to n) 1 / (1 + k/n)
Now, let's consider the interval [0, 1] and partition it into n subintervals of equal width Δx = 1/n. The left endpoint of each subinterval can be represented by x(sub(k)) = k/n, where k ranges from 1 to n.
Now, we can rewrite A(sub(n)) in terms of the Riemann sum:
A(sub(n)) = Σ (k=1 to n) f(x(sub(k))) * Δx
where f(x) = 1 / (1 + x)
Substituting the expression for x(sub(k)), we have:
A(sub(n)) = Σ (k=1 to n) f(k/n) * (1/n)
Now, observe that A(sub(n)) is a right Riemann sum since we are using the right endpoint of each subinterval. To make the connection with the definite integral, we can take the limit as n approaches infinity:
lim(n->∞) A(sub(n)) = lim(n->∞) Σ (k=1 to n) f(k/n) * (1/n)
This limit is equivalent to the definite integral:
∫[0, 1] f(x) dx
where f(x) = 1 / (1 + x).
Evaluating the integral:
∫[0, 1] 1 / (1 + x) dx = ln(1 + x) |[0, 1] = ln(1 + 1) - ln(1 + 0) = ln(2) - ln(1) = ln(2)
Therefore, the limit of A(sub(n)) as n approaches infinity is ln 2.