Question 2
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
a. To show that an > 0 for n ≥ 1, we start by showing that Hn ≥ ln(n+1).
Using the hint provided, let's consider the integral from 1 to n+1 of dx/x:
∫(1 to n+1) dx/x
Taking the integral, we get:
ln(n+1) - ln(1) = ln(n+1)
Now, let's compare this to Hn:
Hn = 1 + 1/2 + .... + 1/n
Notice that Hn has n terms, while the integral has n+1 terms. We can modify the integral to account for this:
ln(n+1) - ln(1) = ln(n+1) - ln(n)
Therefore, for n ≥ 1, we have:
Hn ≥ ln(n+1) ≥ ln(n)
Now, let's define an:
an = Hn - ln(n)
Substituting from the previous inequality, we have:
an ≥ ln(n) - ln(n) = 0
Hence, an > 0 for n ≥ 1.
b. To show that {an} is decreasing, we need to show that an - an+1 ≥ 0.
Notice that an - an+1 can be interpreted as the area between the graphs of an and an+1, both considered as functions of n.
Since we have already established that an > 0 for n ≥ 1, this means that the graph of an is always above the x-axis in the positive region.
Similarly, since an+1 > 0 for n ≥ 1 as well, the graph of an+1 is also always above the x-axis in the positive region.
Therefore, the area between the two graphs is always positive or equal to zero, i.e. an - an+1 ≥ 0.
Thus, {an} is decreasing.
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 suffices to show that {an} is bounded below.
From part (a), we know that an > 0 for n ≥ 1.
Therefore, {an} is bounded below by zero.
Hence, lim n → ∞ an exists as a finite limit since {an} is a decreasing sequence bounded below.
To solve this problem, we'll go step by step to answer each part.
a. To show that an > 0 for n >= 1, we'll use the given hint: show that Hn >= integral from 1 to n+1 dx/x.
The harmonic number Hn is defined as the sum of the reciprocals of positive integers up to n. Therefore, it can be expressed as:
Hn = 1/1 + 1/2 + 1/3 + ... + 1/n
To prove that Hn >= integral from 1 to n+1 dx/x, we can use the fact that the integral of 1/x with respect to x is ln(x).
So, applying the integral from 1 to n+1 dx/x, we get:
∫[1 to n+1] dx/x = ln(n+1) - ln(1) = ln(n+1)
Comparing this integral to Hn, we now need to show that Hn >= ln(n+1).
We can prove this inequality indirectly using induction.
Base case: When n = 1, H1 = 1/1 = 1, and ln(1+1) = ln(2). Since 2 > 1, the inequality holds for the base case.
Inductive step: Assume that Hk >= ln(k+1) for some positive integer k >= 1.
We need to prove that H(k+1) >= ln((k+1)+1), which is equivalent to proving H(k+1) >= ln(k+2).
H(k+1) can be expressed as:
H(k+1) = Hk + 1/(k+1)
By the assumption in the inductive step, Hk >= ln(k+1).
Adding 1/(k+1) to both sides of the inequality, we get:
Hk + 1/(k+1) >= ln(k+1) + 1/(k+1)
H(k+1) >= ln(k+1) + 1/(k+1)
To show that H(k+1) >= ln(k+2), it suffices to prove that ln(k+1) + 1/(k+1) >= ln(k+2).
By the properties of logarithms and algebraic manipulation, we can simplify the right side of the inequality:
ln(k+2) = ln((k+1)+1) = ln(k+1) + ln(1+1/(k+1)) ≥ ln(k+1) + 1/(k+1)
Since H(k+1) >= ln(k+1) + 1/(k+1), the inequality holds for the inductive step.
Therefore, by the principle of mathematical induction, Hn >= ln(n+1) for all n >= 1, which implies that an = Hn - ln(n) > 0 for n >= 1.
b. To show that {an} is decreasing, we can interpret an - an+1 as a positive area.
We know that an - an+1 = (Hn - ln(n)) - (Hn+1 - ln(n+1))
Simplifying this expression, we get:
an - an+1 = ln(n+1) - ln(n) = ln((n+1)/n)
Since the natural logarithm is an increasing function, (n+1)/n > 1 implies ln((n+1)/n) > 0.
Therefore, an - an+1 > 0, meaning that an > an+1. This shows that {an} is decreasing.
c. To prove that lim n -> infinity an exists, we can use the Monotone Convergence Theorem.
We have shown that {an} is a decreasing sequence (from part b). Additionally, an is bounded below by 0 (from part a).
By the Monotone Convergence Theorem, any bounded and decreasing sequence must converge to a limit.
Therefore, lim n -> infinity an exists.