Harrison Chapman

Math 317: Homework 2

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. $$a_n = \frac {7n-19}{3n+7}$$

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

      $$\left\vert \frac {7n-19}{3n+7} - \frac 73 \right\vert = \frac {106}{3(3n + 7)} < \frac {106}{3(3N + 7)} = \epsilon.$$

      So $\lim a_n = 7/3$.

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   2. $$s_n = \frac 1n \sin n$$

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

      $$\left\vert \frac {\sin n}n \right\vert = \frac{\vert \sin n \vert}{\vert n \vert} \le \frac 1{\vert n \vert} = \frac 1n < \frac 1N = \epsilon.$$

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

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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 $\vert s_n \vert < \epsilon$.

      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.

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   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.

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3. (7.4) Give examples of

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

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

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   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 $e$.

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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 $t_n = -s_n$.

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

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

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   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 $s_n = 1/t_n$.

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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 $L < 1$ then $\lim s_n = 0$.

      Proof. As $L < 1$ there exists a number $L < a < 1$ by the Completeness Axiom. Consider $\epsilon = a - L$ which is positive as $L < a$.

      Claim: There exists $N$ such that for all $n \ge N$, $\vert s_{n+1}\vert < a \vert s_n\vert$. Let $N \in \mathbb N$ be that for which

      $$\left\vert \left\vert \frac {s_{n+1}}{s_n} \right\vert - L \right\vert < \epsilon$$

      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

      $$L - \epsilon < \left\vert \frac {s_{n+1}}{s_n} \right\vert < L + \epsilon = a.$$

      Notice that the right part of this inequality implies that $\vert s_{n+1} \vert < a \vert s_n \vert$, as claimed.

      Claim: For all natural numbers $k$, $\vert s_{N+k}\vert < a^k \vert s_N\vert$. We prove this using induction on $k$. The base case of $k=1$ we have already shown, by taking $n = N$ in the inequality $\vert s_{n+1} \vert < a \vert s_n \vert$ which is true by choice of $N$.

      Suppose now that $\vert s_{N+k} \vert < a^k \vert s_N \vert$. We have that $N+k \ge N$, so we know that $\vert s_{(N+k)+1} \vert < a \vert s_{N+k} \vert$. By our inductive hypothesis, we have that

      $$\vert s_{(N+k)+1} \vert < a \vert s_{N+k} \vert < a (a^k \vert s_N \vert) = a^{k+1}\vert s_N \vert,$$

      so by induction the claim is true.

      Claim: $\lim s_n = 0$. Observe that we have now that $0 < \vert s_{N+k} \vert < a^k \vert s_N \vert$. Consider $n > N$; then consider $k = n-N$, which is a natural number as $N \in \mathbb N$ and $n > N$. So

      $$0 < \vert s_{n} \vert < a^n \left\vert \frac {s_N}{a^N} \right\vert$$

      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 $\lim s_n = 0$. $\Box$

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

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

   $$\lim \left\vert \frac{t_{n+1}}{t_n} \right\vert = \lim \left\vert \frac{s_{n}}{s_{n+1}} \right\vert = 1/L < 1.$$

   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 $\lim \vert s_n \vert = +\infty$.

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