Homework 4

Solution.

Let $a = \frac{L + 1}{2} < 1$. Then because $\varepsilon = a - L > 0$, there exists $N$ such that for all $n \geq N$,

(Scratchwork: To see where we get the inductive hypothesis, you can just write out what happens when iterating the inequality. For example, if $n \geq N$, then $n + 1 \geq N$ as well, so

The inductive hypothesis should be clear from here.) We will show that for any $n \geq N$, we have

We proceed by induction on $k$.

Base case: $\p{k = 1}$ This is just ($*$) above.

Inductive step: Assume $\abs{s_{n+k}} < a^k \abs{s_n}$. Then applying ($*$) and the inductive hypothesis gives

In particular, ($\dagger$) implies (by replacing $n$ with $N$ and $k$ with $n - N$)

Note that we need $n > N$ instead of just $n \geq N$ since we need $k \geq 1$ to apply ($\dagger$). Finally, since $\abs{a} < 1$, we can Theorem 9.7(b), so by the squeeze lemma, $\abs{s_n}$ converges and

Here, we used that $N$, and hence $s_N$, are constants. This implies (by a previous homework problem) that $\lim_{n\to\infty} s_n = 0$ as well.

Solution.

1. By the triangle inequality (and technically induction) and the assumption on $\p{s_n}$, we have for $k \geq 0$

   $$\begin{aligned} \abs{s_{n+k} - s_n} = \abs{\sum_{i=n}^{n+k-1} \p{s_{i+1} - s_i}} &\leq \sum_{i=n}^{n+k-1} \abs{s_{i+1} - s_i} \\ &< \sum_{i=n}^{n+k-1} 2^{-i} \\ &< \sum_{i=n}^\infty 2^{-i} \\ &= 2^{-n+1}. \tag{$*$} \end{aligned}$$

   Let $\varepsilon > 0$. Since $2^{-n+1} \to 0$ (by Theorem 9.7(c) with $a = \frac{1}{2}$), there exists $N$ such that if $n \geq N$, then $2^{-n+1} < \varepsilon$. Let $n, m \geq N$. Without loss of generality, we may assume $m \geq n$; in the case $m < n$, we simply swap $m$ and $n$ in the following. By ($*$),

   $$\abs{s_m - s_n} = \abs{s_{n+\p{m-n}} - s_n} < 2^{-n+1} < \varepsilon.$$

2. No. For example, if $s_n = \sum_{i=1}^n \frac{1}{i}$, then

   $$s_{n+1} - s_n = \frac{1}{n+1} < \frac{1}{n},$$

   but $\p{s_n}$ diverges (it's the harmonic series).

Common Mistakes

The only mistake I saw often was in part (2). Basically, if you write something like this:

No, because by the same argument above, we have

$$\abs{s_{n+k} - s_n} < \sum_{i=n}^{n+k-1} \frac{1}{i},$$

and the right-hand side diverges, so the result in (1) does not hold.

While nothing you wrote is incorrect, this argument just doesn't answer the question. All this shows is that our proof of (1) no longer works, but doesn't show that the result is false. For instance, what if there's a better proof that works in both cases that we don't know about?

To show that there's no proof possible, you just need to give a counter-example.

Solution.

Let $n \in \N$. Then because $\sup S$ is the least upper bound of $S$, $\sup S - \frac{1}{n}$ cannot be an upper bound for $S$. Thus, there exists $s_n \in S$ such that

(Note that I used $s_n \leq \sup S$ to get the inequality with an absolute value.) Thus, (by the axiom of countable choice) we get a sequence $\p{s_n}$ of elements in $S$ such that $\abs{s_n - \sup S} \leq \frac{1}{n}$. By the squeeze theorem, we immediately get $s_n \to \sup S$.