Chapter MPS, Problem 11.12MPS

To determine

Prove the conjecture involving $\frac{1}{5}$ .

Answer to Problem 11.12MPS

The expression is the statement to be proved where $n$ has been replaced by $k+1$ .

Explanation of Solution

Given information:

Two statements that can be proved using mathematical induction are, $\begin{array}{l}\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\mathrm{...}+\frac{1}{3^n}=\frac{1}{2}\left(1-\frac{1}{3^n}\right)and\\ \frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\mathrm{...}+\frac{1}{4^n}=\frac{1}{3}\left(1-\frac{1}{4^n}\right).\end{array}$

Calculation:

Consider

  $\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\mathrm{...}+\frac{1}{5^n}=\frac{1}{4}\left(1-\frac{1}{5^n}\right)$

Let $n=1$ , the left side of the equation is $\frac{1}{5}$ .

The right side of equation is,

  $\frac{1}{4}\left(1-\frac{1}{5^1}\right)=\frac{1}{4}\left(\frac{5-1}{5}\right)$

  $\begin{array}{l}=\frac{1}{4}\left(\frac{4}{5}\right)\\ =\frac{1}{5}\end{array}$

Hence, given conjecture is true for $n=1$ .

Assume that the given conjecture is true for $n=k$ .

That is,

  $\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\mathrm{...}+\frac{1}{5^k}=\frac{1}{4}\left(1-\frac{1}{5^k}\right)$ , for a natural number $k$ .

Show that the given conjecture is true for $n=k+1$ .

  $\begin{array}{l}\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\mathrm{...}+\frac{1}{5^k}=\frac{1}{4}\left(1-\frac{1}{5^k}\right)\\ \frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\mathrm{...}+\frac{1}{5^k}+\frac{1}{5^{k+1}}=\frac{1}{4}\left(1+\frac{1}{5^k}\right)+\frac{1}{5^{k=1}}\\ =\frac{1}{4}-\frac{1}{{4.5}^k}+\frac{1}{{5.5}^k}\\ =\frac{1}{4}-\frac{1}{5^k}\left(\frac{1}{4}-\frac{1}{5}\right)\\ =\frac{1}{4}-\frac{1}{5^k}\left(\frac{1}{5.4}\right)\\ =\frac{1}{4}-\frac{1}{4}\left(\frac{1}{5^{k+1}}\right)\\ =\frac{1}{4}\left(1-\frac{1}{5^{k+1}}\right)\end{array}$

Therefore, $\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\mathrm{...}+\frac{1}{5^k}+\frac{1}{5^{k+1}}=\frac{1}{4}\left(1-\frac{1}{5^{k+1}}\right)$

Hence, the above expression is the statement to be proved where $n$ has been replaced by $k+1$ .

Hence, this proves the conjecture.