Calculus Sequences problem…10pts fo best answer?

a) c_1 = 1/1 + 1/2 = 3/2
c_2 = 1/2 + 1/3 + 1/4 = 13/12
c_3 = 1/3 + 1/4 + 1/5 + 1/6 = 19/20

b) Drawing the picture of the rectangles to approximate ∫(n to 2n) dx/x is useful.

Using left endpoints, we have ∫(n to 2n) dx/x < 1/n + 1/(n+1) + ... + 1/(2n-1)
(since the rectangles overshoot the area denoted by the integral).

Using right endpoints, we have ∫(n to 2n) dx/x > 1/(n+1) + 1/(n+2) + ... + 1/(2n)
(since the rectangles undershoot the area denoted by the integral).

Therefore, [∫(n to 2n) dx/x] + 1/(2n) < 1/n + 1/(n+1) + ... + 1/(2n-1) + 1/(2n) = c_n.
and [∫(n to 2n) dx/x] + 1/n > 1/n + 1/(n+1) + 1/(n+2) + ... + 1/(2n) = c_n.

Putting these inequalities together, we have
[∫(n to 2n) dx/x] + 1/(2n) < c_n < [∫(n to 2n) dx/x] + 1/n.

c) Letting n→∞:
lim(n→∞) ∫(n to 2n) dx/x = lim(n→∞) ln x {for x = n to 2n} = ln(2n) - ln n = ln(2n/n) = ln 2.

Hence, lim(n→∞) [∫(n to 2n) dx/x] + 1/(2n) = ln 2 + 0 = ln 2
and lim(n→∞) [∫(n to 2n) dx/x] + 1/n = ln 2 + 0 = ln 2

So, lim(n→∞) c_n = ln 2 by the Squeeze Theorem.

I hope this helps!