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How to calculate $\sum\limits_{n = 1}^\infty{(0.8)^n}$?

I notice that $0.8^n$ is a geometric series with $a = 0.8$ and $r = 0.8$. So $\frac{0.8}{0.2} = 4$ but the answer is $0$?

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  • $\begingroup$ You say "$s_n$ is the sum of the terms from $1$ to $n$", is that just a description, or is it an additional definition of $s_n$? In the first case, we have $\lim_{n\to\infty}(2-3(0.8)^n)=2$, but in the second case, for large $n$, you have $\sum_{n\gt N}^\infty{2}$, which is unbounded... $\endgroup$
    – abiessu
    Commented Sep 27, 2013 at 13:18
  • $\begingroup$ What exactly are you asking? $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:21
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    $\begingroup$ If you want $$\sum_{n=0}^\infty s_n$$ then the series is divergent because of that $\;2\;$ there... $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:26
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    $\begingroup$ With your edits, now you definitely have the value of your sum correct, assuming that you are using $0.8\over 1-0.8$ instead of ${1\over 1-0.8}-1$ due to starting at $n=1$ instead of $n=0$. Perhaps you could write out the entire question including any descriptive paragraphs that go with it? Otherwise, it appears that you are correct and the source of the answer is wrong. $\endgroup$
    – abiessu
    Commented Sep 27, 2013 at 13:35
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    $\begingroup$ I agree with @abiessu; with what you have written in your question, your calculations are correct. This means either the question isn't posted fully or the source you are referring to is incorrect. $\endgroup$
    – Clayton
    Commented Sep 27, 2013 at 13:39

3 Answers 3

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The easiest way to remember it is

$$\mbox{Infinite Sum}= \frac{\mbox{First Term In Series}}{1-\mbox{Common Ratio}}. $$

That way you don't have to worry about whether the series starts with $n=0$, $n=1$, $n=k$, and whether the exponent is $n$, $n+1$, etc. Applying that here we get

$$ \frac{\mbox{First Term In Series}}{1-\mbox{Common Ratio}} =\frac{0.8}{1-0.8}=\frac{0.8}{0.2}=4. $$

EDIT: I just noticed that Don Antonio's comment is essentially the same as my answer, but he writes it in an English sentence.

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Hints:

$$|q|<1\;\implies\;q^n\xrightarrow[n\to\infty]{}0$$

Added: For $\;|q|<1\;$

$$\sum_{n=0}^\infty q^n=\frac1{1-q}$$

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  • $\begingroup$ This doesn't explain why my method is incorrect. $\endgroup$
    – Don Larynx
    Commented Sep 27, 2013 at 13:19
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    $\begingroup$ I've no idea what you do there: why and what for do you do the division $\;\frac{0.8}{0.2}\;$ ? What do you think this is giving you? $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:20
  • $\begingroup$ Also, you still haven't answered why I have 4 and the answer is $0$. a/(1-r) = 0.8/(1-0.8). $\endgroup$
    – Don Larynx
    Commented Sep 27, 2013 at 13:34
  • $\begingroup$ because only you, perhaps, can understand why to divide $\;0.8/0.2\;$ and what this has to do with your question: I do not understand why you did that. $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:35
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    $\begingroup$ Well, now that you've finally wrote the correct question then yes: the sum of an infinite descending geometric series equals the first summand divided by one minus the common quotient, so the answer indeed is 4 $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:40
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There are a couple typical formulas for sums over geometric series as follows for $|q|\lt 1$:

$$\sum_{n=0}^\infty q^n={1\over 1-q}$$

$$\sum_{n=1}^\infty q^n={q\over 1-q}={1\over 1-q}-1$$

The far RHS of the second formula simply notes the difference between the two formulas due to the initial term of $q^0=1$ not being a part of the sum.

In particular, you have $\sum_{n=1}^\infty 0.8^n={0.8\over 1-0.8}=4$, just as you have calculated.

EDIT:

As noted by DonAntonio, the general formula takes the following form for $k\ge 0$:

$$\sum_{n=k}^\infty q^n={q^k\over 1-q}$$

In fact, some quick mental estimations suggest that for $q\ge 0$, $k\in \mathbb Z$ also works, but that extension would need to be verified and clearly does not cover $-1\lt q \lt 0$.

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    $\begingroup$ Or in general $$\sum_{n=k}^\infty q^n=\frac{q^k}{1-q}\;\;,\;\;|q|<1$$ $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:50
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    $\begingroup$ @DonAntonio: FWIW, I think your answer should stand since it was first, and answers both the original question as asked and the question in its current form. $\endgroup$
    – abiessu
    Commented Sep 27, 2013 at 13:51
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    $\begingroup$ my question stands, don't worry. What answer to choose is up to the OP. $\endgroup$
    – DonAntonio
    Commented Sep 27, 2013 at 13:52

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