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I have to solve - $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^i$$ - geometric series?

The geometric series sequence I know is - $$\sum_{i=0}^\infty x_i= \frac{1}{1-x}$$

However in my assignment, the series starts from $i=1$.

The solution I have is - $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^i = \frac{1}{1-\frac{5}{12}}-1$$

Can you explain please why is that the solution?

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    $\begingroup$ Hint: Write out the first few terms. $\endgroup$ – Botond Jan 16 at 13:57
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HINT: $$\sum_{i=0}^\infty x_i= \frac{1}{1-x} =x_0 + \sum_{i=1}^\infty x_i$$

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    $\begingroup$ which means $\frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks. $\endgroup$ – Alan Jan 16 at 14:01
  • $\begingroup$ exactly correct @Alan $\endgroup$ – Ahmad Bazzi Jan 16 at 14:01
  • $\begingroup$ Thank you for your answer and for your time. $\endgroup$ – Alan Jan 16 at 14:02
  • $\begingroup$ no problem @Alan .. you can mark the answer as correct if you found it useful :) $\endgroup$ – Ahmad Bazzi Jan 16 at 14:03
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    $\begingroup$ Yes, in a few minutes when I can :-) $\endgroup$ – Alan Jan 16 at 14:03
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There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series: $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}=x\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i-1}= x \sum_{i=0}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}$$ (setting $\;i\leftarrow i-1$), so $$\sum_{i=1}^\infty \left(\frac{5}{12}\right)^{\mkern-5mu i}=\frac x{1-x}.$$

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  • $\begingroup$ Thank you, a great point of view. $\endgroup$ – Alan Jan 16 at 14:23
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Observe that for suitable $x$: $$(1-x)\sum_{i=k}^{\infty}x^i=\sum_{i=k}^{\infty}x^i-\sum_{i=k+1}^{\infty}x^i=x^k$$so that:$$\sum_{i=k}^{\infty}x^i=\frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=\frac5{12}$.

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