Question

If $r < 1$ and positive and m is a positive integer, show that $\left( 2m+1 \right){{r}^{m}}\left( 1-r \right) < 1-{{r}^{2m+1}}$. Hence show that $n{{r}^{n}}$ is infinitely small when n is infinitely great.

Answer 1

We have to form a G.P. of $a,ar,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},........,a{{r}^{n-1}}$. Then we need to apply the theorem of $A.M. > G.M.$ taking dual terms from both ends together like $1,{{r}^{2m}}$, $r,{{r}^{2m-1}}$………
We get the inequality of $\left( 2m+1 \right){{r}^{m}} < \dfrac{1-{{r}^{2m+1}}}{1-r}$ and find the proof for $\left( 2m+1 \right){{r}^{m}}\left( 1-r \right) < 1-{{r}^{2m+1}}$.

Complete step by step answer:
We try to form the given inequality in the form of a series of G.P.
In $\left( 2m+1 \right){{r}^{m}}\left( 1-r \right) < 1-{{r}^{2m+1}}$, we divide by $\left( 1-r \right)$.

& \left( 2m+1 \right){{r}^{m}}\left( 1-r \right) < 1-{{r}^{2m+1}} \\\ & \Rightarrow \dfrac{\left( 2m+1 \right){{r}^{m}}\left( 1-r \right)}{\left( 1-r \right)} < \dfrac{1-{{r}^{2m+1}}}{\left( 1-r \right)} \\\ & \Rightarrow \left( 2m+1 \right){{r}^{m}} < \dfrac{1-{{r}^{2m+1}}}{\left( 1-r \right)} \\\ \end{aligned}$$ The right part is a sum of a G.P. series. So, we take a G.P. series of $1,r,{{r}^{2}},{{r}^{3}},{{r}^{4}},........,{{r}^{2m}}$ where $r < 1$ and positive and m is a positive integer. So, in a series of G.P. of $a,ar,a{{r}^{2}},a{{r}^{3}},a{{r}^{4}},........,a{{r}^{n-1}}$ where $r < 1$ the sum will be $$a\dfrac{1-{{r}^{n}}}{1-r}$$. Here, $a=1,n=2m$. So, sum of $1,r,{{r}^{2}},{{r}^{3}},{{r}^{4}},........,{{r}^{2m}}$ will be $$\dfrac{1-{{r}^{2m+1}}}{1-r}$$. Now we have the middle term of the series $1,r,{{r}^{2}},{{r}^{3}},{{r}^{4}},........,{{r}^{2m}}$ as ${{r}^{m}}$. There are $\left( 2m+1 \right)$ terms in the series $1,r,{{r}^{2}},{{r}^{3}},{{r}^{4}},........,{{r}^{2m}}$. We have the theorem of $A.M. > G.M.$ In the series $1,r,{{r}^{2}},{{r}^{3}},{{r}^{4}},........,{{r}^{2m}}$, we take dual terms from both ends together like $1,{{r}^{2m}}$, $r,{{r}^{2m-1}}$…….. and apply the above theorem of $A.M. > G.M.$ So, $\dfrac{1+{{r}^{2m}}}{2} > \sqrt{1\times {{r}^{2m}}}\Rightarrow 1+{{r}^{2m}} > 2{{r}^{m}}$ and $\dfrac{r+{{r}^{2m-1}}}{2} > \sqrt{r\times {{r}^{2m-1}}}\Rightarrow r+{{r}^{2m-1}} > 2{{r}^{m}}$ This process continues for all the terms of the series $1,r,{{r}^{2}},{{r}^{3}},{{r}^{4}},........,{{r}^{2m}}$ other than the middle term ${{r}^{m}}$as there are $\left( 2m+1 \right)$ terms in the series. We take the sum of the inequalities $\begin{aligned} & \left( 1+{{r}^{2m}} \right)+\left( r+{{r}^{2m-1}} \right)+......+\left( {{r}^{m-1}}+{{r}^{m+1}} \right) > 2{{r}^{m}}+2{{r}^{m}}+...... \\\ & \Rightarrow 1+{{r}^{2m}}+r+{{r}^{2m-1}}+.....+{{r}^{m-1}}+{{r}^{m+1}} > 2m{{r}^{m}} \\\ & \Rightarrow 1+r+{{r}^{2}}+.....+{{r}^{m-1}}+{{r}^{m+1}}+........+{{r}^{2m}} > 2m{{r}^{m}} \\\ \end{aligned}$. We add the middle term on the both sides ${{r}^{m}}$. $\begin{aligned} & 1+r+{{r}^{2}}+.....+{{r}^{m-1}}+{{r}^{m+1}}+........+{{r}^{2m}} > 2m{{r}^{m}} \\\ & \Rightarrow 1+r+{{r}^{2}}+.....+{{r}^{m-1}}+{{r}^{m}}+{{r}^{m+1}}+........+{{r}^{2m}} > 2m{{r}^{m}}+{{r}^{m}} \\\ & \Rightarrow 1+r+{{r}^{2}}+.....+{{r}^{m-1}}+{{r}^{m}}+{{r}^{m+1}}+........+{{r}^{2m}} > \left( 2m+1 \right){{r}^{m}} \\\ & \Rightarrow \left( 2m+1 \right){{r}^{m}} < 1+r+{{r}^{2}}+.....+{{r}^{m-1}}+{{r}^{m}}+{{r}^{m+1}}+........+{{r}^{2m}} \\\ \end{aligned}$ We have the sum of the left side of the inequality where $$1+r+{{r}^{2}}+........+{{r}^{2m}}=\dfrac{1-{{r}^{2m+1}}}{1-r}$$. We replace the value and get $\begin{aligned} & \left( 2m+1 \right){{r}^{m}} < 1+r+{{r}^{2}}+.....+{{r}^{m-1}}+{{r}^{m}}+{{r}^{m+1}}+........+{{r}^{2m}} \\\ & \Rightarrow \left( 2m+1 \right){{r}^{m}} < \dfrac{1-{{r}^{2m+1}}}{1-r} \\\ & \Rightarrow \left( 2m+1 \right){{r}^{m}}\left( 1-r \right) < 1-{{r}^{2m+1}} \\\ \end{aligned}$ Thus proved. Now $r < 1$ and positive and n is a positive integer. In the range of $-1 < r < 1$ we have the curve of $n{{r}^{n}}$ as tending to 0. As $r < 1$, it’s more like a fraction and we can express it as $\dfrac{1}{m}$ where $m > 1$. So, ${{r}^{n}}=\dfrac{1}{{{m}^{n}}}$. Here, $n\to \infty \Rightarrow {{m}^{n}}\to \infty \Rightarrow \left( {{r}^{n}}=\dfrac{1}{{{m}^{n}}} \right)\to 0$. **So, $n{{r}^{n}}$ is infinitely small when n is infinitely great.** **Note:** We need to be careful when we are trying to take duals as there are an odd number of terms in the series. The term n in $n{{r}^{n}}$ will not be considered at the time of finding the value of $n{{r}^{n}}$. Even though n tends to infinity the value of ${{r}^{n}}$ steers the rate of the value as it has the indices.