# Help finishing proof via induction for a summation

So I have to prove the following equation using induction for n >= 2: $$\sum\limits_{i=1}^n 4/5^i < 1$$ However the question asks me to prove something stronger such as this: $$\sum\limits_{i=1}^n 4/5^i <= 1 - \frac{1}{5^n}$$ first to imply the first equation is true. So far I have the following:

Base Case: Let n = 2

$$\sum\limits_{i=1}^2 4/5^i = \frac{4}{5} + \frac{4}{25} = \frac {24}{25}$$ then I also applied it to $$1 - \frac{1}{5^n} \rightarrow 1 - \frac{1}{5^2} = \frac{24}{25}$$ Therefore I can make the following assumptions yes?

Inductive Hypothesis for all 2 <= n <= k it is $$\sum\limits_{i=1}^n 4/5^i = 4\frac{\frac{1}{5^n} - 1}{\frac{1}{5} - 1} = 1 - \frac{1}{5^n} < 1$$ Inductive Step Hopefully I'm ok up to here, I'll show what I have so far for this step. $$\sum\limits_{i=1}^{k+1} 4/5^i = \frac{\frac{1}{5^{k+1}} - 1}{\frac{1}{5} - 1} = 4\frac{(\frac{1}{5^k}-1) * \frac{1}{5} - \frac{4}{5}}{\frac{1}{5} -1}$$ $$= \frac{1}{5} * 4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1} - 4\frac{\frac{4}{5}}{\frac{1}{5} - 1}$$ so here I have: $$4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1}$$ which I know is: $$= \sum\limits_{i=1}^k 4/5^i$$ which is my inductive hypothesis, I am unsure of how to finish my proof from here... any help correcting or finishing the proof is very much appreciated

## 2 Answers

Hint: $$\sum_{i=1}^{k+1}4/5^i=\sum_{i=1}^k 4/5^i+4/5^{k+1}$$ Use the induction hypothesis on the sum from $1$ to $k$ and simplify.

$$\sum_{i=1}^{3}1/5^i=(1/5+1/5^2+1/5^3)*(1-1/5)/(1-1/5)=(1-1/5^3)/4$$ By induction it is easy to see that

$$\sum_{i=1}^{n}1/5^i = (1-1/5^n)/4$$

And so

$$\sum_{i=1}^{n}4/5^i = (1-1/5^n)$$

Which is strictly less than one.