Chapter 10.4, Problem 6QR

To determine

To find: The smallest positive integer N for series ${\displaystyle \sum a_n}$ and ${\displaystyle \sum b_n}$ for which $a_n>b_n,\forall n\ge N$ .

Answer to Problem 6QR

For series ${\displaystyle \sum a_n}={\displaystyle \sum n^2}$ and ${\displaystyle \sum b_n}={\displaystyle \sum 5n}$ , there exist smallest positive integer $N=5$ such that $a_n>b_n,\forall n\ge N$ .

Explanation of Solution

Given information:

The given series are ${\displaystyle \sum a_n}={\displaystyle \sum 5n}$ and ${\displaystyle \sum b_n}={{\displaystyle \sum n}}^2$ or ${\displaystyle \sum a_n}={{\displaystyle \sum n}}^2$ and ${\displaystyle \sum b_n}={\displaystyle \sum 5n}$

Formula used:

The terms of one series is always greater, equal of less than another series.

Calculations:

Case I: The given series are ${\displaystyle \sum a_n}={\displaystyle \sum 5n}\Rightarrow a_n=5n$ and ${\displaystyle \sum b_n}={{\displaystyle \sum n}}^2\Rightarrow b_n=n^2$ .

Now, $a_n>b_n,\forall n\ge N$

  $\Rightarrow 5n>n^2,\forall n\ge N$

  $\Rightarrow 5n-n^2>0,\forall n\ge N$

  $\Rightarrow n\left(5-n\right)>0,\forall n\ge N$

  $\Rightarrow n\ge 0and\left(5-n\right)>0,\forall n\ge N$

  $\Rightarrow n\ge 0and5>n,\forall n\ge N$

Thus, for $N=0$ , $a_n>b_n,\forall n\ge N$ where ${\displaystyle \sum a_n}={\displaystyle \sum 5n}$ and ${\displaystyle \sum b_n}={{\displaystyle \sum n}}^2$ .

Case II: The given series are ${\displaystyle \sum a_n}={\displaystyle \sum n^2}\Rightarrow a_n=n^2$ and ${\displaystyle \sum b_n}={\displaystyle \sum 5n}\Rightarrow b_n=5n$ .

Now, $a_n>b_n,\forall n\ge N$

  $\Rightarrow n^2>5n,\forall n\ge N$

  $\Rightarrow n^2-5n>0,\forall n\ge N$

  $\Rightarrow n\left(n-5\right)>0,\forall n\ge N$

  $\Rightarrow n\ge 0andn\ge 5,\forall n\ge N$

Thus, for $N=0,N=5$ , $a_n>b_n,\forall n\ge N$ for series ${\displaystyle \sum a_n}={\displaystyle \sum n^2}$ and ${\displaystyle \sum b_n}={\displaystyle \sum 5n}$ .

Conclusion:

For series ${\displaystyle \sum a_n}={\displaystyle \sum n^2}$ and ${\displaystyle \sum b_n}={\displaystyle \sum 5n}$ , there exist smallest positive integer $N=5$ such that $a_n>b_n,\forall n\ge N$ .