## Wednesday, November 1, 2006

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