**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 $

it seems like the first one should converge, since its partial sums are just a subset of the partial sums of the original sequence, which converge.

ReplyDeletefor the second one you can probably find some wacky counterexample by making the signs all weird huh?

Yup. The example for the second one would be

ReplyDeletea_n = ((-1)^n)/n,

and the terms of the new sequence would all have absolute value > 1/4.