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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

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2 Answers 2

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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.

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$$\sum_{i=1}^{3}1/5^i=(1/5+1/5^2+1/5^3)*(1-1/5)/(1-1/5)=(1-1/5^3)/4 $$ 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.

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