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Find the value of $c$ for which the series equals the indicated sum. $$ \sum \limits_{n=0}^{\infty} e^{cn} = 5$$

The $n=0$ made me think of geometric series.

So I solve it like this:

$r = e^{c}$ thus $ 5 = \frac{1}{1-e^{c}}$

resulting in $5e^c = 4$ which gives you $c = \ln \frac{4}{5}$

is this any alternative method for solving this?

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    $\begingroup$ That’s certainly how I’d solve it. $\endgroup$ – Brian M. Scott Mar 3 '13 at 13:11
  • $\begingroup$ @BrianM.Scott is it possible to solve in a different manner? $\endgroup$ – yiyi Mar 3 '13 at 13:13
  • $\begingroup$ Probably, but I can’t think of another at the moment. $\endgroup$ – Brian M. Scott Mar 3 '13 at 13:15
  • $\begingroup$ @GitGud I have tried putting the value of $c$ in to the sum on my calculator, but it doesn't compute a result. So, I was asking if there was another method to check my results. $\endgroup$ – yiyi Mar 3 '13 at 13:28
  • $\begingroup$ @MaoYiyi Are you looking to check if your solution is right or are you looking for alternative ways to solve this? In order to check if your value for $c$ satisfies the equality, note that $$e^{cn}=e^{nc}=e^{n\ln {4/5}}=e^{\ln {(4/5)^n}}$$ and recall that $\exp$ and $\ln$ are inverse functions of each other. $\endgroup$ – Git Gud Mar 3 '13 at 13:31
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$$ 5 = \sum_{n=0}^\infty e^{cn} = 1 + \sum_{n=1}^\infty e^{cn} = 1 + e^{c}\sum_{n=1}^\infty e^{c(n-1)} = 1 + e^c \sum_{n=0}^\infty e^{cn} = 1+5e^c $$ so $$ e^c = \frac{4}{5} $$ and $c=\log(4/5)$.

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