Proving That $\sum^{n}_{k=0} \bigl(\frac{4}{5}\bigr)^k < 5$

Question

Using induction, prove that $$\sum_{k=0}^n \biggl(\frac 4 5 \biggr)^k = 1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n<5$$ for all natural numbers $n.$

What I have tried is as follows.

Consider the statement $$P(n):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^n<5.$$

For $n=1$, we have that $\displaystyle P(1):1+\frac{4}{5}<5$ is true.

For $n=k$, we assume that $$\displaystyle P(k):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^k<5.$$

$\displaystyle P(k+1):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^k+\bigg(\frac{4}{5}\bigg)^{k+1}<5+\bigg(\frac{4}{5}\bigg)^{k+1}$

How can I prove that the sum on the left is $< 5?$ Help me please. Thanks.

Answer 1

Assume $$ 1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^k < 5 $$ for some positive integer $k$.$\;$Then \begin{align*} & \frac{4}{5} {\,\cdot} \left( 1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^k \right) < \frac{4}{5}{\,\cdot\,}5 \\[4pt] \implies\;& \frac{4}{5}+\left(\frac{4}{5}\right)^2+\left(\frac{4}{5}\right)^3+\cdots +\left(\frac{4}{5}\right)^{k+1} < 4 \\[4pt] \implies\;& 1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^{k+1} < 5 \\[4pt] \end{align*} which completes the induction.

Answer 2

$$1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n = 5\left(1-\dfrac{4^{n+1}}{5^{n+1}} \right)$$