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Homework Statement
Let y_{n} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} - \int_{1}^{n} \frac{1}{t}dt Prove that the sequence \{y_{n}\}_{i=1}^{n} converges
Homework Equations
The Attempt at a Solution
y_{1} = 1
y_{2} = 1 +\frac{1}{2} - \int_{1}^{2} \frac{1}{t}dt = 1 + \frac{1}{2} - ln2
y_{3} = 1 +\frac{1}{2} + \frac{1}{3} - \int_{1}^{3} \frac{1}{t}dt = \frac{11}{6} - ln3
Now we can see that in a base case y_n is decreasing. If I can finish induction (and then show it is bounded, I can show it is convergent.
So then Assume y_n < y_n-1 to show y_n+1 < y_n.
I get a really nasty inequality that I am going to try to type in, and this is where I get stuck on this problem.
y_n+1 = 1 + 1/2 + ... + 1/n + 1/(n+1) - [ \int_{1}^{n} \frac{1}{t}dt + \int_{n}^{n+1} \frac{1}{t} dt ]
= 1+1/2+...+1/n+1/(n+1) - ln(n) - ln(n+1) + ln(n)
< [1+ 1/2 +...+ 1/n - ln(n)] + 1/(n+1) - ln(n+1) + ln(n) and by the induction hypothesis we get
< [1 + 1/2 + ... + 1/(n-1) - ln(n-1)] + 1/(n+1) - ln(n+1) + ln(n) and obviously
< 1+ 1/2 + ... + 1/(n-1) + 1/n - ln(n-1) - ln(n+1) + ln(n)
= 1 + 1/2 + ... + 1/n - ln(n+1) %comment(ln((n^2+n)/n) = ln(n+1))
< 1 + 1/2 + ... + 1/n - ln(n) = y_n
So I get what I wanted, and I don't think I made any mistakes, but this is about as ugly as it gets. Does anyone have any more elegant ways to do what my goal is (see way up there some where showing {y_n} converges
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