Induction summation proof: $\sum_{i=1}^{n} \frac{4}{5^{i}} < 1$

Question

Don't want a full answer but can somebody help me in the right direction with this problem. Have to prove using induction

$$\forall n \geqslant 2: \sum_{i=1}^{n} \frac{4}{5^{i}} < 1$$

Answer 1

You can find an explicit formula like this $$\begin{align}S&=\text{}\frac4{5}+&&\!\!\!\!\!\!\!\!\!\!\!\!\frac4{5^2}+\ldots+\frac4{5^n}\\ \frac15S&=&&\!\!\!\!\!\!\!\!\!\!\!\!\frac4{5^2}+\ldots+\frac4{5^3}+\frac4{5^{n+1}}\end{align}$$

Therefore $\displaystyle S-\frac15S=\frac45-\frac4{5^{n+1}}$ and $\displaystyle S=\frac{\frac45-\frac4{5^{n+1}}}{1-\frac15}=1-\frac1{5^n}<1$.

Now you can prove the $\displaystyle S=1-\frac1{5^n}$ by induction :)

Answer 2

It's rather silly to use induction on this one (all you need is the sum of a geometric series), but if you insist: if $S_n$ is your sum, express $S_{n+1}$ as $a + S_n/5$ for suitable $a$.

Answer 3

Show that the difference between 1 and the previous sum is always greater than next term.

$$ Assume: 1 - \sum_1^{n - 1}\frac{4}{5^i} > \frac{4}{5^n} $$

If this is true? Is it true for the next case well? $$ next-case:1 - \sum_1^n \frac{4}{5^i} \stackrel{?}{>} \frac{4}{5^{n + 1}} $$