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I am trying to find the sum of the series,

$\large\Sigma_{n=0}^\infty \large\frac{(-1)^n}{5^n}$

and I have no idea even how to start.

The only way I know to find sums is:
1)By geometric series $\large r^n$ where $r < 1$.
2)By partial sums, which I also tried but was unsuccessful, as I was not able to make a formula from the sums.

I can see that the sum is geometric with $\large r=\frac{1}{5}$, but I have no idea how to attain the answer $\large-\frac{1}{6}$
(which seems to be correct according to WolframAlpha) with the formula $\large S_n=\large\frac {1}{1-r}$.

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  • $\begingroup$ 1) works fine here, $r = -\frac{1}{5}$. But the sum is $\frac{5}{6}$. It would be $-\frac{1}{6}$ if the sum started at $n = 1$. $\endgroup$ – Daniel Fischer Nov 20 '14 at 14:37
  • $\begingroup$ Thank you, Daniel. I see now that WolframAlpha automatically bounds the series from 1 to infinity, not from 0. Is it much harder to calculate the sum from 1? $\endgroup$ – Akitirija Nov 20 '14 at 14:46
  • $\begingroup$ @Akitirija Note that $\frac{(-1)^0}{5^0}=1$. You only need to subtract it to $\frac{5}{6}$. $\endgroup$ – Vincenzo Oliva Nov 20 '14 at 14:49
  • $\begingroup$ No, not at all. Method 1: subtract $1$. method 2: multiply with $r$. $\endgroup$ – Daniel Fischer Nov 20 '14 at 14:49
  • $\begingroup$ Ah! Very cool! Thank you again, Daniel! $\endgroup$ – Akitirija Nov 20 '14 at 14:54
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If you also want to know where does the sum formula comes,

Consider a series with first term=$a$ and common ratio=$r$
Let,
$S=a+a.r+a.r^2+a.r^3+...+\infty$ $\;\;\;\;\;\;\;$ (i)

Multiplying whole series by $r$ we get,

$r.S=a.r+a.r^2+a.r^3+...+\infty$ $\;\;\;\;\;\;\;\;$ (ii)

Subtracting (ii) from (i)

$(1-r).S=a$ $\;\;\;\;\;$ $\implies$ $S=\large\frac{a}{1-r}$

$\large\Sigma_{n=0}^\infty \large\frac{(-1)^n}{5^n}=1-\large\frac{1}{5}+\large\frac{1}{25}+..+upto\space infinite\space terms$
From series $a=1$
$\;\;\;\;\;\;\;\;\;\;\;$ and $r=-1/5$

$S=\large\frac{1}{1-\large\frac{-1}{5}}$ $\;\;\;\;\;\;$ $\implies$ $S=\large\frac{5}{6}$

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  • $\begingroup$ Very cool, thank you, Mann! $\endgroup$ – Akitirija Nov 20 '14 at 15:40
  • $\begingroup$ No problem, glad to help! :) $\endgroup$ – Mann Nov 20 '14 at 15:42
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This is a geometric series with $r=-1/5$. The sum is $1/(1-(-1/5))=5/6$.

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$$\sum_{n=0}^{\infty}\frac{(-1)^n}{5^n}=\sum_{n=0}^{\infty}\big(\frac{-1}{5}\big)^n$$

So $a=1$ and $r=-1/5$.

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  • $\begingroup$ Thank you, dleggas. I see now that WolframAlpha automatically bounds the series from 1 to infinity, not from 0. So the correct answer is 5/6. $\endgroup$ – Akitirija Nov 20 '14 at 14:42

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