91Ó°ÊÓ

Given sequence is:

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

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.

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$\underset{k→∞}{\lim }\frac{k}{k+1}=1$.

Given sequence is:

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

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 $0.01$.

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

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

$k≈-11.51$and $k≈8.51$

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

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

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