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Using induction, verify that the following equation is true for every positive integer $n$. $$r^0 + r^1 +\ldots+ r^n < \frac{1}{1−r},$$ for all $n ≥0$, and $0 < r < 1$.

I am completely stuck on how to approach this problem after the base case has been proven. Can anyone help me out?

Edit. I apologize for not explaining more. I did not understand that we could still use $r$ as a constant in this case to solve for $n$. I was of the understanding that I would also have to prove $r$ using induction and did not understand how that would be possible while proving $n$.

Gary
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OldNoodleCup
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Feb 16 '23 at 02:20
  • Use the formula for the sum of a finite geometric sequence on the left-hand side. – Gary Feb 16 '23 at 02:22
  • Use the finite sum of geometric sequence. When you do that, try dividing it into two parts and notice something about it. – Hamdiken Feb 16 '23 at 02:33
  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Christian E. Ramirez Feb 16 '23 at 03:17
  • @Gary I apologize for not adding more. My confusion stemmed from the fact that I did not realize we could treat r as a constant and just solve for n. I was trying to figure out how to solve for both r and k simultaneously. My future posts will include more detail. – OldNoodleCup Feb 16 '23 at 04:08
  • @C-RAM I apologize for not adding more. My confusion stemmed from the fact that I did not realize we could treat r as a constant and just solve for n. I was trying to figure out how to solve for both r and k simultaneously. My future posts will include more detail. – OldNoodleCup Feb 16 '23 at 04:10

1 Answers1

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Note that $$ r^0 + r^1 + \ldots + r^{k} + r^{k+1} = 1 + r\cdot(r^0 + r^1 + \ldots + r^{k}). $$ So if the inequality is true for $n=k$, then $$ r^0 + r^1 + \ldots + r^{k} + r^{k+1} < 1 + \frac{r}{1-r}=\frac{1}{1-r}; $$ that is, it's also true for $n=k+1$.

mjqxxxx
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