W07 Regular

Due date: Sunday 3/8, 11:59pm

Sequences

01

03

Computing the terms of a sequence

Calculate the first four terms of each sequence from the given general term, starting at $n=1$:

(a) $\cos \pi n$ $$ (b) ${\displaystyle \frac{n!}{2^n}}$ $$ (c) ${(-1)}^{n+1}$ $$ (d) ${\displaystyle \frac{n}{n+1}}$ $$ (e) ${\displaystyle \frac{3^n}{n!}}$ $$ (f) ${\displaystyle \frac{(2n-1)!}{n!}}$

Link to original

Solution

Solutions - 0140-03

(a) $-1,+1,-1,+1$

(b) ${\displaystyle \frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{3}{2}}$

(c) $+1,-1,+1,-1$

(d) ${\displaystyle \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}}$

(e) ${\displaystyle 3,\frac{9}{2},\frac{9}{2},\frac{27}{8}}$

(f) $1,3,20,210$

Link to original

02 $★$

02

Squeeze theorem

Determine whether the sequence converges, and if it does, find its limit:

(a) $a_n={\displaystyle \frac{{\cos }^{2}n}{2^n}}$ $$ (b) $b_n={\displaystyle {(2^n+3^n)}^{1/n}}$

(Hint for (b): Verify that $3\le b_n\le {(2\cdot 3^n)}^{1/n}$.)

Link to original

Solution

Solutions - 0140-02

(a)

(1) Set up squeeze relations:

$$0\le \frac{{\cos }^{2}n}{2^n}\le \frac{1}{2^n}$$

---

(2) Apply theorem:

We have:

$$\underset{n\to \infty }{\lim }\frac{1}{2^n}=0$$

Therefore:

$$\underset{n\to \infty }{\lim }\frac{{\cos }^{2}n}{2^n}=0$$

We conclude that $a_n$ converges.

---

(b)

(1) Generate squeeze inequalities:

Observe:

$$3^n\le 2^n+3^n\le 3^n+3^n$$

Rewrite RHS:

$$3^n+3^n\gg \gg 2{(3)}^n$$

Raise all terms to $1/n$:

$$3\le {(2^n+3^n)}^{1/n}\le (2^{1/n})3$$

---

(2) Apply squeeze theorem:

$$2^{1/n}\to 1\text{as}n\to \infty$$

Therefore:

$$(2^{1/n})3\to 3\text{as}n\to \infty$$

Conclude that: ${(2^n+3^n)}^{1/n}\to 3$

Link to original

Series

03 $★$

05

Series from its partial sums

Suppose we know that the partial sums $S_N$ of a series $S={\sum }_{n=1}^{\infty }{a}_n$ are given by the formula $S_N=5-\frac{2}{N^2}$.

(a) Compute $a_3$.

(b) Find a formula for the general term $a_n$.

(c) Find the sum $S$.

Link to original

Solution

Solutions - 0150-05

(a)

$$\begin{array}{c}{a}_3=S_3-S_2\gg \gg (5-\frac{2}{3^2})-(5-\frac{2}{2^2})\\ \\ \gg \gg -\frac{2}{9}+\frac{2}{4}\gg \gg \frac{5}{18}\end{array}$$

---

(b)

$$\begin{array}{c}{a}_n=S_n-S_{n-1}\gg \gg (5-\frac{2}{n^2})-(5-\frac{2}{{(n-1)}^2})\\ \\ \gg \gg \frac{2}{{(n-1)}^2}-\frac{2}{n^2}\\ \\ \gg \gg \frac{2n^2-2{(n-1)}^2}{n^2{(n-1)}^2}\\ \\ \gg \gg \frac{2n^2-2(n^2-2n+1)}{n^2{(n-1)}^2}\\ \\ \gg \gg \frac{4n-2}{n^2{(n-1)}^2}\end{array}$$

When $n=1$, this formula is undefined, because $S_0$ is undefined. But we know that:

$$a_1=S_1\gg \gg 5-\frac{2}{1^2}\gg \gg 3$$

---

(c)

Simply take the limit of $S_N$ as $N\to \infty$:

$$\begin{array}{c}{S}_N\stackrel{n\to \infty }{⟶}S\\ \\ 5-\frac{2}{N^2}\stackrel{n\to \infty }{⟶}5\end{array}$$Link to original

04 $★$

06

Partial sums and total sum

Consider the series:

$$\sum _{n=1}^{\infty }\frac{{(-8)}^{n-1}}{9^n}$$

(a) Compute a formula for the $N^{\text{th}}$ partial sum $S_N$ by applying the “shift method” steps using the values in this series.

(b) By taking the limit of this formula as $N\to \infty$, find the value of the series.

(c) Find the same value of the series by computing $a_0$ and $r$ and plugging into $S=\frac{a_0}{1-r}$.

Link to original

Solution

Solutions - 0150-06

(a) (1) Rewrite general term in standard form for a geometric series:

$$S_N=\sum _{n=0}^N\left(\frac{1}{9}\right){(-\frac{8}{9})}^n$$

Thus $a_0=1/9$ and $r=-8/9$.

(2) Apply “shift method” technique:

Compare $S_N$ and $r{S}_N$:

$$\begin{array}{cc}{S}_N& =\frac{1}{9}+\left(\frac{1}{9}\right){(-\frac{8}{9})}^1+\left(\frac{1}{9}\right){(-\frac{8}{9})}^2+\cdots +\left(\frac{1}{9}\right){(-\frac{8}{9})}^N\\ & \\ (-\frac{8}{9})S_N& =\left(\frac{1}{9}\right){(-\frac{8}{9})}^1+\left(\frac{1}{9}\right){(-\frac{8}{9})}^2+\cdots +\left(\frac{1}{9}\right){(-\frac{8}{9})}^N+\left(\frac{1}{9}\right){(-\frac{8}{9})}^{N+1}\end{array}$$

Subtract and cancel terms:

$$S_N-(-\frac{8}{9})S_N=\frac{1}{9}-\left(\frac{1}{9}\right){(-\frac{8}{9})}^{N+1}$$

Factor and solve:

$$\begin{array}{c}{S}_N(1+\frac{8}{9})=\frac{1}{9}(1-{(-\frac{8}{9})}^{N+1})\\ \\ \gg \gg S_N=\frac{9}{17}\cdot \frac{1}{9}(1-{(-\frac{8}{9})}^{N+1})\\ \\ \gg \gg S_N=\frac{1}{17}(1-{(-\frac{8}{9})}^{N+1})\end{array}$$

---

(b) In the limit as $N\to \infty$, this converges to $1/17$ because the term ${(-8/9)}^{N+1}$ converges to $0$.

---

(c)

$$S=\frac{1/9}{1--8/9}\gg \gg \frac{1}{17}$$Link to original