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I have to show that $\sum_{i=1}^{\infty} \frac{A}{(1+r)^i} = \frac{A}{r}$. Where $ -1 < r < 1$. I tried putting it like: $$\sum_{i=1}^{\infty} \frac{A}{(1+r)^i} = \frac{A}{r+1} + \sum_{i=2}^{\infty} \frac{A}{(1+r)^i}$$ However I don't know how to continue from here on. This is not a duplicate, most other questions require $|1+r|>1$ which is not necessarily the case here. I would be very grateful if you could help me out or give me pointers how to solve this!

Jan
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  • See https://en.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series – lab bhattacharjee Sep 21 '15 at 07:55
  • Defining $x=\frac 1 {1+r}$ would make life easier (I guess). – Claude Leibovici Sep 21 '15 at 07:59
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    That requires $|1+r|$ to be smaller than 1, which is not always the case here. – Jan Sep 21 '15 at 08:01
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    "This is not a duplicate, most other questions require |1+r|>1 which is not necessarily the case here." Indeed, the fact that you are not requiring $|1+r|>1$ makes the question absurd rather than a dup. Say $r=-1/2$, then $1/(1+r)^i=2^i$ and what would be the sum of the series $\sum A2^i$? – Did Sep 21 '15 at 08:12

2 Answers2

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Hint:

Multiply the sum by $\dfrac1{1+r}$. You get

$$\sum_{i=1}^{\infty} \frac{A}{(1+r)^{i+1}}=\sum_{i=2}^{\infty} \frac{A}{(1+r)^i}. $$

Then add $\dfrac A{1+r}$, and you are back to where you started from

$$\sum_{i=1}^{\infty} \frac{A}{(1+r)^i}.$$

So if the series converges $$\frac S{1+r}+\frac A{1+r}=S.$$

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What if you put out the $A$ term of your serie? Indeed, it doesn't depend on $i$. Thus, you may recognize a convergent serie as $\frac{1}{(1+r)^i} < 1$. What do you know about the value of $\sum q^i$ where $q < 1$?

yarmenti
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