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I need to calculate $$\sum_{n=1}^{\infty} \left(\frac{3}{4}\right)^{n}$$

The sum of geometric series, if $|α|<1$ is $\frac{1}{1-α}$

So my solution was $\frac{1}{1 - \frac{3}{4}}$ but apparently the correct one is $\frac{\frac{3}{4}}{1 - \frac{3}{4}}$

Can you explain why this is the case? It has to do with starting from $n=1$ instead of $0$?

5xum
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Than1
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    "It has to do with starting from $n=1$ instead of $0$?" Yes. Recognize that $\sum\limits_{n=0}^\infty a^n = 1 + \sum\limits_{n=1}^\infty a^n$ and so starting from zero and starting from one are not equal, so of course the result if it started from zero is different than the result if started from one. Add $1$ to your result and it will be made correct. – JMoravitz Oct 08 '21 at 12:28
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    A recent, similar question – ACB Oct 08 '21 at 12:29
  • You recieved 2 answers to your question. Is any of them what you needed? If so, consider accepting the best answer and upvoting all useful answers you got. That's how the site works. – 5xum Oct 11 '21 at 09:25
  • @5xum what do you mean by accepting? I always upvote the useful answers but i dont know what you mean by accepting – Than1 Oct 11 '21 at 22:02
  • @Mosies You can accept an answer by clicking on the check mark beside the answer to toggle it from greyed out to filled in. See here: https://stackoverflow.com/help/someone-answers – 5xum Oct 12 '21 at 07:16
  • @Mosies Any reason why none of the answers is accepted? What is still missing? – 5xum Oct 15 '21 at 10:29

2 Answers2

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Note that it starts at $n=1$, your formula only holds for the case where the sum starts at $n=0$.

spinosarus123
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The sum of geometric series, if $|α|<1$ is $\frac{1}{1-α}$

Yes, but that's only true for the geometric series starting at $0$. In other words, if $|\alpha|<1$, then

$$\sum_{n=0}^\infty \alpha^n = \frac{1}{1-\alpha}$$

You are calculating $$\sum_{n=1}^\infty \alpha^n$$ which is almost what you need. Can you think of any connection between the two expressions?

5xum
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  • I tried to find what we get if n was 0, which was 1. Then i thought i may substract it from the general sum. Is this correct? The result checks out at least – Than1 Oct 08 '21 at 12:43
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    @Mosies Indeed, you can use the fact that, for any infinite sum, you have $$\sum_{n=0}^\infty a_n = a_0 + \sum_{n=1}^\infty a_n$$ and this means that for your particular case, you have $$\sum_{n=0}^\infty \alpha^n = 1 + \sum_{n=1}^\infty \alpha^n$$ because $a_n=\alpha^n$ in your case, and $a_0=\alpha^0=1$.

    But there's another way, specific to geometric series. And that is, to notice that

    $$\sum_{n=1}^\infty\alpha^n = \sum_{n=1}^\infty (\alpha\cdot \alpha^{n-1}) = \alpha\cdot\sum_{n=1}^\infty \alpha^{n-1}$$

     – 5xum Oct 08 '21 at 12:47
  • i see, thanks a lot :) – Than1 Oct 08 '21 at 12:50
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    Now, after relabelling (for example, setting $m=n-1$, you can see that

    $$\sum_{n=1}^\infty\alpha^{n-1} = \sum_{m=0}^\infty \alpha^m$$

    so you can conclude that $$\sum_{n=1}^\infty \alpha^n =\alpha \sum_{n=0}^\infty \alpha^n = \frac{\alpha}{1-\alpha}$$

     – 5xum Oct 08 '21 at 12:50