The Cheat Is To The Limit. Topic: Calculus.

Problem: (1997 IMC Day 1 - #1) Let $ \{e_n\}^{\infty}_{n=1} $ be a sequence of positive reals with $ \displaystyle \lim_{n \rightarrow \infty} e_n = 0 $. Find

$ \displaystyle \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln{\left(\frac{k}{n}+e_n\right)} $.

Solution: First of all, you should notice that this looks a lot like a Reimann Sum, so maybe we can transform it into an integral. In fact, this will solve the problem. First, bound the limit from below by

$ \displaystyle \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln{\left(\frac{k}{n}+e_n\right)} \ge \int^1_0 \ln{x}dx = [x\ln{x}-x]^1_0 = -1 $.

We then also have

$ \displaystyle \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln{\left(\frac{k}{n}+e_n\right)} \le \int_0^1 \ln{(x+e)} dx = [(x+e)\ln{(x+e)}-(x+e)]^1_0 = (1+e)\ln(1+e)-(1+e) $

since for $ n $ big enough that we have $ e_n \le e $. Thus we can take $ e \rightarrow 0 $ as $ e_n \rightarrow 0 $. This means

$ \displaystyle -1 \le \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \ln{\left(\frac{k}{n}+e_n\right)} \le -1+(1+e)\ln(1+e)-e $.

Since the upper bound and lower bound both converge to $ -1 $, the limit is $ -1 $. QED.

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Comment: A classic example of Reimann Sums, which every calculus student should be familiar with (otherwise they aren't really learning calculus...). Applying some simple bounding techniques gives us the answer pretty easily.

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Practice Problem: (1997 IMC Day 1 - #2) Let $ a_n $ be a sequence of reals. Suppose $ \displaystyle \sum a_n $ converges. Do these sums converge as well?

(a) $ a_1+a_2+(a_4+a_3)+(a_8+\cdots+a_5)+(a_{16}+\cdots+a_9)+\cdots $

(b) $ {a_1+a_2+(a_3)+(a_4)+(a_5+a_7)+(a_6+a_8)+(a_9+a_{11}+a_{13}+a_{15}) +(a_{10}+a_{12}+a_{14}+a_{16})+\cdots $