W07 Stepwise

Due date: Friday 2/26, 11:59pm

Sequences

01

01

L’Hopital practice - converting indeterminate form

By imitating the technique of the L’Hopital’s Rule example, find the limit of the sequence:

$$a_n=\sqrt{n}\ln (1+\frac{1}{n})$$

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Solution

Solutions - 0140-01

(1) Indeterminate form:

$$a_n\to \infty \cdot 0\text{as}n\to \infty$$

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(2) L’Hopital:

Convert:

$$a_n=\sqrt{n}\ln (1+\frac{1}{n})\gg \gg \frac{\ln (1+1/n)}{n^{-1/2}}\to \frac{0}{0}$$

Change to $x$ and apply L’Hopital:

$$\frac{\ln (1+1/x)}{x^{-1/2}}\stackrel{d/dx}{⟶}\frac{\frac{-1/x^2}{1+1/x}}{-1/2\cdot x^{-3/2}}\gg \gg \frac{2}{x^{+1/2}+x^{-1/2}}$$

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(3) Take limit:

$$\frac{2}{x^{-1/2}+x^{+1/2}}⟶0\text{as}x\to \infty$$

Therefore $a_n\to 0$ as $n\to \infty$.

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02

05

Limits and convergence

For each sequence, either write the limit value (if it converges), or write ‘diverges’.

(a) ${\displaystyle \frac{5n-1}{12n+9}}$ $$ (b) ${\displaystyle {(-1)}^n\left(\frac{5n-1}{12n+9}\right)}$ $$ (c) ${\displaystyle \sqrt{4+\frac{1}{n}}}$

(d) ${\cos }^{-1}\left({\displaystyle \frac{n^3}{n^3+1}}\right)$ $$ (e) ${\displaystyle 10+{(-\frac{1}{9})}^n}$ $$

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Solution

Solutions - 0140-05

(a) $5/12$ $$ (b) diverges $$ (c) $2$

(d) $0$

Observe that $\frac{n^3}{n^3+1}\to 1$ as $n\to \infty$, but for each $n$, the value is below $1$, in the domain of ${\cos }^{-1}(x)$, which is continuous for $x\in (-1,+1)$.

(e) $10$

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03

04

General term of a sequence

Find a formula for the general term (the $n^{\text{th}}$ term) of each sequence:

(a) ${\displaystyle \frac{1}{1},\frac{-1}{8},\frac{1}{27},\dots }$ $$ (b) ${\displaystyle \frac{2}{6},\frac{3}{7},\frac{4}{8},\dots }$ $$ (c) ${\displaystyle \frac{3}{5},-\frac{4}{25},\frac{5}{125},-\frac{6}{625},\dots }$

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Solution

Solutions - 0140-04

(a) ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{1}{n^3}}$

(b) ${\displaystyle \sum _{n=1}^{\infty }\frac{n+1}{n+5}}$

(c) ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{n+2}{5^n}}$

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Series

04

01

General term of a series

Write this series in summation notation:

$$\frac{1}{1}-\frac{2^2}{2\cdot 1}+\frac{3^3}{3\cdot 2\cdot 1}-\frac{4^4}{4\cdot 3\cdot 2\cdot 1}+\cdots$$

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

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Solution

Solutions - 0150-01

$$\sum _{n=1}^{\infty }{(-1)}^{n-1}\frac{n^n}{n!}$$Link to original