Let an = Hn - ln(n), where Hn is the nth harmonic number
Hn = 1 + 1/2 + .... + 1/n
a. Show that an > 0 for n>= 1 (Hint: show Hn >= integral from 1 to n+1 dx/x)
b. Show {an} is decreasing by interpreting an - an+1 as a positive area. (Hint: use the previous part)
c. Prove lim n-> infinity an exists
Any help is much appreciated. Thank you!!
a. To show that an > 0 for n ≥ 1, we can use the given hint and show that Hn ≥ ∫(1 to n+1) dx/x.
The harmonic number Hn can be written as the sum of integrals:
Hn = ∫(1 to 2) dx/x + ∫(2 to 3) dx/x + ... + ∫(n to n+1) dx/x
Now, observe that ∫(k to k+1) dx/x = ln(k+1) - ln(k), where ln denotes the natural logarithm. Therefore, we have:
Hn = [ln(2) - ln(1)] + [ln(3) - ln(2)] + ... + [ln(n+1) - ln(n)]
Simplifying the expression, we get:
Hn = ln(n+1) - ln(1)
Hn = ln(n+1)
Now, we need to compare this to the integral ∫(1 to n+1) dx/x. The integral can be evaluated as follows:
∫(1 to n+1) dx/x = ln(n+1) - ln(1)
Since ln(n+1) > ln(1) = 0, we've shown that Hn > ∫(1 to n+1) dx/x, which implies that an = Hn - ln(n) > 0 for n ≥ 1.
b. To show that {an} is decreasing, we can consider the expression an - an+1 and interpret it as a positive area.
Starting with an = Hn - ln(n) and an+1 = Hn+1 - ln(n+1), we can write:
an - an+1 = (Hn - ln(n)) - (Hn+1 - ln(n+1))
= (Hn - Hn+1) + (ln(n+1) - ln(n))
= -Hn+1 + ln(n+1) - ln(n)
We have already shown that Hn = ln(n+1), so we can rewrite the expression as:
an - an+1 = -Hn+1 + ln(n+1) - ln(n)
= -ln(n+1) + ln(n+1) - ln(n)
= -ln(n)
Since ln(n) > 0 for n ≥ 1, we conclude that an > an+1, meaning that {an} is a decreasing sequence.
c. To prove that lim n→∞ an exists, we need to show that {an} is a bounded sequence. Since we have already shown that {an} is decreasing, it is sufficient to show that {an} has a lower bound.
We know that an > 0 for n ≥ 1, so the sequence is bounded below by zero. Therefore, since {an} is a decreasing sequence that is bounded below, it must converge. Thus, lim n→∞ an exists.