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Homework 8

Mos Kullathon
921425216

1. Find the following limits

a. $\lim_{n\to\infty}\frac{(-1)^{n^3}}{n}$

By squeeze theorem:
$$\lim_{n\to\infty}-\frac{1}{n} \le \lim_{n\to\infty}\frac{(-1)^{n^3}}{n} \le \lim_{n\to\infty}\frac{1}{n} \\ 0 \le \lim_{n\to\infty}\frac{(-1)^{n^3}}{n} \le 0 \\ \therefore \lim_{n\to\infty}\frac{(-1)^{n^3}}{n}=0.$$

b. $\lim_{n\to\infty}\frac{n^2+1}{n+100}$

$$\deg(n^2+1) > \deg(n+100) \\ \therefore \lim_{n\to\infty}\frac{n^2+1}{n+100} = \infty$$

c. $\lim_{n\to\infty}(\frac{1}{5e})^n$

$$\lim_{n\to\infty}\(\frac{1}{5e}\)^n = \lim_{n\to\infty}\frac{1^n}{5e^n} = \frac{\displaystyle\lim_{n\to\infty}1^n}{\displaystyle\lim_{n\to\infty}5e^n} = 0$$

d. $\lim_{n\to\infty}\frac{n^{100}}{e^{0.01n}}$

The polynomial $n^{100}$ grows slower than the exponential $e^{0.01n}$.

$$\therefore \lim_{n\to\infty}\frac{n^{100}}{e^{0.01n}} = 0$$

e. $\lim_{n\to\infty}(1+\frac{1}{n})(2+\frac{\cos n}{3n^2})$

$$\begin{align*} &\lim_{n\to\infty}\(1+\frac{1}{n}\) \(2+\frac{\cos n}{3n^2}\) \\ &= \underbrace{\lim_{n\to\infty}\(1+\frac{1}{n}\)} _{1} \lim_{n\to\infty}\(2+\frac{\cos n}{3n^2}\) \\ &= \lim_{n\to\infty}2+ \lim_{n\to\infty}\frac{\cos n}{3n^2} \end{align*}$$

By squeeze theorem:
$$\lim_{n\to\infty}-\frac{1}{3n^2} \leq \lim_{n\to\infty}\frac{\cos n}{3n^2} \leq \lim_{n\to\infty}\frac{1}{3n^2} \\ 0 \leq \lim_{n\to\infty}\frac{\cos n}{3n^2} \leq 0 \\ \therefore \lim_{n\to\infty}\frac{\cos n}{3n^2}=0.$$

As such,
$$\begin{align*} &\lim_{n\to\infty}\(1+\frac{1}{n}\) \(2+\frac{\cos n}{3n^2}\) \\ &= \lim_{n\to\infty}2+0 \\ &= 2. \end{align*}$$

f. $\lim_{n\to\infty}\frac{\ln\sqrt{n+1}}{n^2}$

$$\begin{align*} \lim_{n\to\infty}\frac{\ln\sqrt{n+1}}{n^2} &= \lim_{n\to\infty}\frac{\frac{1}{2}\ln(n+1)}{n^2} \\ &= \lim_{n\to\infty}\frac{\frac{1}{2(n+1)}}{2n} \\ &= \lim_{n\to\infty}\frac{1}{4n(n+1)} \\ &= 0 \end{align*}$$

g. $\lim_{n\to\infty}\frac{n!}{n^n}$

$n^n$ grows faster than $n!$.
$$\therefore\lim_{n\to\infty}\frac{n!}{n^n} =0$$

h. $\lim_{n\to\infty}2^{1/n}$

$$\lim_{n\to\infty}2^{1/n} = 2^{\lim_{n\to\infty}1/n} = 2^0 = 1$$

i. $\lim_{n\to\infty}2^{n/(n^2+1)}$

$$\lim_{n\to\infty}2^{n/(n^2+1)} = 2^{\lim_{n\to\infty}n/(n^2+1)} =2^0=1$$

j. $\lim_{n\to\infty}n^{1/n}$

$$\lim_{n\to\infty}n^{1/n} = \lim_{n\to\infty} e^{\frac{1}{n}\ln n} = e^{\lim_{n\to\infty}\frac{\ln n}{n}} = e^0 = 1$$

k. $\lim_{n\to\infty}(-1)^n \tan\frac{n}{n^5-1}$

By squeeze theorem:
$$\lim_{n\to\infty}-1\cdot \tan\frac{n}{n^5-1} \leq \lim_{n\to\infty}(-1)^n\tan\frac{n}{n^5-1} \leq \lim_{n\to\infty}1\cdot\tan\frac{n}{n^5-1} \\ -1(0) \leq\lim_{n\to\infty}(-1)^n\tan\frac{n}{n^5-1}\leq 1(0) \\ 0\leq\lim_{n\to\infty}(-1)^n\tan\frac{n}{n^5-1}\leq 0 \\ \therefore \lim_{n\to\infty}(-1)^n\tan\frac{n}{n^5-1} = 0.$$

2. Evaluate the following infinite series

(i) $\sum_{n=1}^\infty(\frac{2}{5})^n$

The common ratio $r$ is
$$r=\frac{\(\frac{2}{5}\)^2}{\frac{2}{5}} =\frac{\(\frac{2}{5}\)^3}{\(\frac{2}{5}\)^2} =\frac{\(\frac{2}{5}\)^n}{\(\frac{2}{5}\)^{n-1}} =\frac{2}{5}.$$
Since $|r|<1$, the series converges.

Where the initial term $a=\frac{2}{5}$, using the geometric sum formula yields:

$$\begin{align*} \sum_{n=1}^\infty\(\frac{2}{5}\)^n&= \frac{\frac{2}{5}}{1-\frac{2}{5}} \\ &= \frac{2}{3} \end{align*}$$

(ii) $\sum_{n=2}^\infty(-1)^n\frac{5}{7^n}$

The common ratio $r$ is
$$r=\frac{\frac{5(-1)^3}{7^3}}{\frac{5(-1)^2}{7^2}} =\frac{\frac{5(-1)^4}{7^4}}{\frac{5(-1)^3}{7^3}} =\frac{\frac{5(-1)^n}{7^n}}{\frac{5(-1)^{n-1}}{7^{n-1}}} =-\frac{1}{7}.$$

Since $|r|<1$, the series converges.

Where the initial term $a=\frac{5(-1)^2}{7^2}=\frac{5}{49}$, using the geometric sum formula yields:

$$\begin{align*} \sum_{n=2}^\infty(-1)^n\frac{5}{7^n} &= \frac{\frac{5}{49}}{1+\frac{1}{7}} \\ &= \frac{5}{56} \end{align*}$$

(iii) $4 + \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots$

$$\begin{align*} &4 + \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots \\ &= 4 + \sum_{n=1}^\infty\frac{3}{10^n} \end{align*}$$

For $\displaystyle\sum_{n=1}^\infty\frac{3}{10^n}$:

The common ratio $r$ is
$$r=\frac{\frac{3}{10^2}}{\frac{3}{10}} =\frac{\frac{3}{10^3}}{\frac{3}{10^2}} =\frac{\frac{3}{10^n}}{\frac{3}{10^{n-1}}} =\frac{1}{10}$$

Since $|r|<1$, the series converges.

Where the initial term $a=\frac{3}{10}$, using the geometric sum formula yields:
$$\begin{align*} \sum_{n=1}^\infty\frac{3}{10^n} &= \frac{\frac{3}{10}}{1-\frac{1}{10}} \\ &= \frac{1}{3}. \\ \therefore 4+\sum_{n=1}^\infty\frac{3}{10^n} &= 4+\frac{1}{3} \\ &= \frac{13}{3} \end{align*}$$

(iv) $\sum_{n=1}^\infty(\ln2)^n,\quad\sum_{n=1}^\infty(\ln3)^n$ (Be careful of the convergence)

For $\displaystyle\sum_{n=1}^\infty(\ln2)^n$:

The common ratio $r$ is
$$r=\frac{\ln^22}{\ln2}=\frac{\ln^32}{\ln^22} =\frac{\ln^n2}{\ln^{n-1}2} = \ln2 \approx 0.69.$$

Since $|r|<1$, the series converges.

Where the initial term $a=\ln2$, using the geometric sum formula yields:
$$\begin{align*} \sum_{n=1}^\infty(\ln2)^n &= \frac{\ln2}{1-\ln2} \approx 2.26. \end{align*}$$

For $\displaystyle\sum_{n=1}^\infty(\ln3)^n$:

The common ratio $r$ is
$$r=\frac{\ln^23}{\ln2}=\frac{\ln^33}{\ln^23} =\frac{\ln^n3}{\ln^{n-1}3} = \ln3 \approx 1.10.$$

Since $|r|>1$, the series diverges.