Question

Q16. If r < 1 and positive and m is a positive integer, show that $\sf{(2m+1)r^m(1-r)<1-r^{(2m+1)}}$. And also show that, nrⁿ is indefinitely small when n is indefinitely great.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

r < 1

It means 1 - r > 0

So,

$\rm :\longmapsto\: {(1 - {r}^{m} )}^{2} > 0$

$\rm :\longmapsto\:1 + {r}^{2m} - 2 {r}^{m} > 0$

$\bf\implies \:1 + {r}^{2m} > 2 {r}^{m}$

can be rewritten as

$\rm :\longmapsto\:1 + {r}^{2m} > 2 \sqrt{{r}^{2m} \times 1}$

$\rm :\implies\:AM > GM$

Let consider Arithmetic mean and Geometric mean relationship between

$\rm :\longmapsto\:r \: and \: {r}^{2m - 1}$

$\rm :\longmapsto\:r + {r}^{2m - 1} \geqslant 2 \sqrt{r \times {r}^{2m - 1} }$

$\rm :\longmapsto\:r + {r}^{2m - 1} \geqslant 2 \sqrt{{r}^{2m - 1 + 1} }$

$\rm :\longmapsto\:r + {r}^{2m - 1} \geqslant 2 \sqrt{{r}^{2m} }$

$\rm :\longmapsto\:r + {r}^{2m - 1} \geqslant 2{r}^{m} - - - (2)$

Let consider Arithmetic mean and Geometric mean relationship between

$\rm :\longmapsto\: {r}^{2} \: and \: {r}^{2m - 2}$

$\rm :\longmapsto\: {r}^{2} + {r}^{2m - 2} \geqslant 2 \sqrt{ {r}^{2} \times {r}^{2m - 2} }$

$\rm :\longmapsto\: {r}^{2} + {r}^{2m - 1} \geqslant 2 \sqrt{{r}^{2m - 2 + 2} }$

$\rm :\longmapsto\: {r}^{2} + {r}^{2m - 1} \geqslant 2 \sqrt{{r}^{2m} }$

$\rm :\longmapsto\: {r}^{2} + {r}^{2m - 1} \geqslant 2 {r}^{m} - - - (3)$

and

so on

Let consider Arithmetic mean and Geometric mean relationship between

$\rm :\longmapsto\:{r}^{m} \: and \: {r}^{m}$

$\rm :\longmapsto\:{r}^{m } + {r}^{m} \geqslant 2 \sqrt{{r}^{2m} }$

$\rm :\longmapsto\:{r}^{m } + {r}^{m} \geqslant 2 {r}^{m} - - - (4)$

On adding these m + 1 equations, we get

$\rm \: (1 + r + {r}^{2} + - - + {r}^{2m}) + {r}^{m} > 2(m + 1){r}^{m}$

$\rm \: (1 + r + {r}^{2} + - - + {r}^{2m}) > 2(m + 1){r}^{m} - {r}^{m}$

$\rm \: (1 + r + {r}^{2} + - - + {r}^{2m}) > (2m + 2){r}^{m} - {r}^{m}$

$\rm \: (1 + r + {r}^{2} + - - + {r}^{2m}) > (2m + 2 - 1){r}^{m}$

$\rm \: (1 + r + {r}^{2} + - - + {r}^{2m}) > (2m + 1){r}^{m}$

Left hand is a GP series consisting of 2m + 1 terms having first term unity and common ratio r < 1.

So, LHS can be represented by sum of 2m + 1 terms as

$\rm :\longmapsto\:\dfrac{1(1 - {r}^{2m + 1} )}{1 - r} > (2m + 1){r}^{m}$

$\rm :\longmapsto\:\dfrac{1 - {r}^{2m + 1} }{1 - r} > (2m + 1){r}^{m}$

$\rm :\longmapsto\:1 - {r}^{2m + 1} > (2m + 1){r}^{m}(1 - r)$

$\bf\implies \:\: (2m + 1){r}^{m}(1 - r) < 1 - {r}^{2m + 1}$

Hence, Proved

Now, we further know that

$\rm :\longmapsto\:x \in \: (0,1) \: then \: {x}^{n} \to \: 0 \: if \: n \: is \: very \: large$

In above relationship, that we prove

Let n = m

$\bf\implies \:\: (2n + 1){r}^{n}(1 - r) < 1 - {r}^{2n + 1}$

$\bf\implies \:\: (2n + 1){r}^{n} <( 1 - {r}^{2n + 1}) \div (1 - r)$

$\bf\implies \:\: 2n {r}^{n} + {r}^{n} <( 1 - {r}^{2n + 1}) \div (1 - r)$

$\bf\implies \:\: 2n {r}^{n} <( 1 - {r}^{2n + 1}) \div (1 - r) \: - \: {r}^{n}$

$\bf\implies \:\: 2n {r}^{n} <1$

$\bf\implies \:\: n {r}^{n} < \dfrac{1}{2}$

Hence,

$\bf\implies \: {nr}^{n} \: is \: very \: small$