Harrison Chapman

Math 317: Homework 2

Due: Friday, February 8, 2019
1. (8.2 b,e) For each of the following sequences, determine its limit (usual Calculus reasoning will be helpful here), then prove that the sequence does indeed converge to this limit.

1. Let $\epsilon > 0$. Consider $N = \frac 16 \left( \frac {106}{3\epsilon} - 7 \right)$. Then for $n > N$,

So $\lim a_n = 7/3$.

2. Let $\epsilon > 0$. Consider $N = 1/\epsilon$ and $n > N$. Then

So $\lim (\sin n)/n = 0$.

2. (8.6) Let $(s_n)$ be a sequence in $\mathbb R$.

1. Prove that $\lim{s_n} = 0$ if and only if $\lim{ | s_n | } = 0$. (Hint: This is an “if and only if” statement, so you will need to prove both directions of the statement.)

($\Rightarrow$) Suppose $\lim s_n = 0$. Consider $\epsilon > 0$. Then there exists $N$ so that $n > N$ implies that $% $.

Observe that $\vert \vert s_n \vert - 0 \vert = \vert s_n \vert$ which is less than $\epsilon$ by above. So $\lim \vert s_n \vert = 0$.

($\Leftarrow$) The reverse direction is very similar to the forward direction.

2. Prove that if $s_n = (-1)^n$ that $\lim{|s_n|}$ exists but $s_n$ diverges.

We have that $s_n$ diverges by Example 4 in Section 8 (pp41-42). Also, $\vert s_n \vert = 1$ which converges to $1$ as it is constant.

3. (7.4) Give examples of

1. A sequence of irrational numbers converging to a rational number

For example, let $s_n = \pi/2^n$.

2. A sequence of rational numbers converging to an irrational number

For example, let $s_n = \sum_{k=0}^n{1/(k!)}$ which converges to $e$.

4. For each of the following false statements, give a counterexample.

1. If $s_n$ and $t_n$ are two divergent sequences, their sum $s_n + t_n$ also diverges.

For example, let $s_n$ diverge and let $t_n = -s_n$.

2. If a sequence $s_n$ converges, so too does the sequence $t_n = \sum_{i=1}^n{s_i}$.

For example, let $s_n = 1/n$.

3. If the sequence $s_n$ converges and the sequence $t_n$ diverges, the product $s_nt_n$ also diverges.

For example, let $t_n$ be any divergent sequence and let $s_n = 0$.

Another example, let $t_n$ be any divergent, nonzero sequence and let $s_n = 1/t_n$.

5. (9.12) Suppose $s_n \ne 0$ for all $n \in \mathbb N$ and that the limit $\lim {\left|\frac{s_{n+1}}{s_n}\right|} = L$ exists.

1. Show that if $% $ then $\lim s_n = 0$.

Proof. As $% $ there exists a number $% $ by the Completeness Axiom. Consider $\epsilon = a - L$ which is positive as $% $.

Claim: There exists $N$ such that for all $n \ge N$, $% $. Let $N \in \mathbb N$ be that for which

for all $n \ge N$, which exists as $\lim {\left\vert \frac{s_{n+1}}{s_n}\right\vert} = L$. (Notice: For simplicity’s sake, we’re taking $N$ here so that even $n = N$ satisfies the convergence inequality. We may do this, as once we know $N$ with the strict inequality exists, we can just choose the next natural number larger by the Archimedean Property.) So, for such $n$ we have that

Notice that the right part of this inequality implies that $% $, as claimed.

Claim: For all natural numbers $k$, $% $. We prove this using induction on $k$. The base case of $k=1$ we have already shown, by taking $n = N$ in the inequality $% $ which is true by choice of $N$.

Suppose now that $% $. We have that $N+k \ge N$, so we know that $% $. By our inductive hypothesis, we have that

so by induction the claim is true.

Claim: $\lim s_n = 0$. Observe that we have now that $% $. Consider $n > N$; then consider $k = n-N$, which is a natural number as $N \in \mathbb N$ and $n > N$. So

Notice we’re actually considering the sequence $(\vert s_n \vert)_{n=N}^{\infty}$ here. But think about why if we can prove the limit for this sequence, we’ve also proved the limit for the full sequence as well.

The LHS has limit $0$, and the RHS has limit $0$ (why?). So by the squeeze theorem, the sequence $(\vert s_n \vert)_{n=N}^{\infty}$ has limit 0 and so $\lim \vert s_n \vert = 0$ as well. By question (2a), this means $\lim s_n = 0$. $\Box$

2. Show that if $L > 1$ then $\lim | s_n | = +\infty$.

Consider the sequence $t_n = 1/s_n$. So

So $\lim t_n = 0$ and by question 2a, $\lim \vert t_n \vert = 0$. As $\vert s_n \vert$ is a positive sequence and $\lim 1/\vert s_n \vert = \lim \vert t_n \vert = 0$ we have that $\lim \vert s_n \vert = +\infty$.