# Let $(L_n)$ be a sequence defined by $(1/(n+1)) + (1/(n+1)) + … 1/(2n))$. Show that $(L_n)$ is monotonic increasing.

I am having difficulty understanding the sequence in the following question (Exercise 2.26 out of Real Analysis by Howie).

Context question:

Let $$(L_n)$$ be a sequence defined by $$\frac{1}{n+1} + \frac{1}{n+1} + ...+ \frac{1}{2n}.$$ Show that $$(L_n)$$ is monotonic increasing, has 1 as an upper bound, and deduce that $$L_n$$ has limit $$L$$, where $$1/2 \leq L \leq 1$$.

[In fact, as we shall see in Chapter 6 the limit is $$log (2)$$].

$$L_{n+1}-L_n = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} \geq \frac{1}{2n+2} + \frac{1}{2n} - \frac{1}{n+1} = 0.$$

My question:

I do not understand what the sequence $$L_n$$ looks like or how it behaves. Specifically, I do not understand how to interpret the fact that the first two terms are given by $$\frac{1}{n+1}$$ but the final term is given by $$\frac{1}{2n}$$.

The answer begins with a familiar strategy: subtract the first term from the second term and see if it's positive. However, I'm still not sure what the given sequence even really looks like, since it's not like anything I've ever encountered before or when searching online.

There is a clue in the answer that suggests that the last term $$\frac{1}{2n}$$ is correct. Specifically, this suggests a calculation like: $$(\frac{1}{n} + \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}) -(\frac{1}{n} + \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})$$ where each term of the sequence is determined by the series $$\sum \frac{1}{2n}$$ familiar from Calculus 2.

However, there are two reasons why this cannot be what is meant. First, it was not in the errata for the question or the solution. Second, a computation $$\sum \frac{1}{2n+2} - \sum \frac{1}{2n}$$ would not have the term $$-\frac{1}{n+1}$$ left over.

This is the extent of my understanding, and I would hate to move on to the next section leaving any questions unanswered.

Just to clarify, I am not looking for a complete solution to the problem, just clarification about the given sequence. Thank you!

• I suspect that the second term is $1/(n+2)$, not a duplicate of the first term. So there's a typo, maybe yours, maybe the book's. – Ethan Bolker May 29 at 22:02
• Its the book's typo, let me try that and see, since that would make more sense, and potentially satisfy the bound of 1. Update: While I am not done, I'm convinced that's what it was. I will close – fantasticasm89 May 29 at 22:02

Update, thank you to Ethan Bolker for suggesting that the second term is $$\frac{1}{n+2}$$.