Q 8. - Calculus

Question

Consider the sequence $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},...,\frac{k}{k+1},....$.

(a) What happens to the terms of this sequence as $k$ gets larger and larger? Express your answer in limit notation.

(b) Use a calculator to find a sufficiently larger value of $k$ so that every term past the $kth$ term of this sequence will be within $0.01$ unit of $1$. 

Step-by-Step Solution

Answer

Part (a) The limit notation is $\underset{k\to \infty }{\lim }\frac{k}{k+1}=1$.

Part (b) The sufficiently large value of $k$ is $k=9$.

Part (a) Step 1. Given information.

Given sequence is:

$\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},...,\frac{k}{k+1},....$

Part (a) Step 2. Find the terms of this sequence as k gets larger and larger in limit notation.

The $k^{th}$ term is:

$f\left(k\right)=\frac{k}{k+1}$

Make the table for the value of $\frac{k}{k+1}$ when the value of $k$ gets larger and larger.

k	0	25	50	100	1000	10000
$\frac{k}{k+1}$	1	0.9615	0.9803	0.99	0.999	0.9999

From the above table as $k$ gets larger and larger, the quantity $\frac{k}{k+1}$ approaches to $1$.

So, the required answer in limit notation is $\underset{k\to \infty }{\lim }\frac{k}{k+1}=1$.

Part (b) Step 1. Given information.

Given sequence is:

$\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},...,\frac{k}{k+1},....$

Part (b) Step 2. Find a sufficiently large value of k .

First write the $k^{th}$ term and ${\left(k+1\right)}^{th}$ term:

$\begin{array}{rcl}f\left(k\right)& =& \frac{k}{k+1}\\ f(k+1)& =& \frac{k+1}{k+2}\end{array}$

It is given that every term past the $k^{th}$ of this sequence will be with in $0.01$.

$\begin{array}{rcl}f(k+1)-f\left(k\right)& <& 0.01\\ \frac{k+1}{k+2}-\frac{k}{k+1}& <& 0.01\\ \frac{k^2+1+2k-k^2-2k}{(k+2)(k+1)}& <& 0.01\\ \frac{1}{(k+2)(k+1)}& <& 0.01\\ \frac{1}{0.01}& <& (k+2)(k+1)\\ 100& <& k^2+3k+2\end{array}$

Part (b) Step 3. Continue the above step.

$\begin{array}{rcl}{k}^2+3k+2& =& 100\\ k^2+3k-98& =& 0\\ k& =& \frac{-3\pm \sqrt{3^2-4\times 1\times \left(-98\right)}}{2\times 1}\\ & =& \frac{-3\pm \sqrt{9+392}}{2}\\ & =& \frac{-3\pm \sqrt{401}}{2}\\ k& \approx & \frac{-3\pm 20.02}{2}\end{array}$

$k\approx -11.51$ and $k\approx 8.51$

For every $k>9$, the terms will be:

For $k=9$, $\frac{10}{11}-\frac{9}{10}\approx 0.009$

Hence, the sufficiently large value of $k$ is $k=9$.