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Find the value of $c$ such that $$\sum_{n=0}^\infty e^{nc}=16$$

I am not too confident with the direction to pursue, but so far, I have let $$a(n)=e^{nc}$$ I then calculated $\dfrac{a(n+1)}{a(n)}$, to which I got $e^c$ I tried to find the radius of convergence by solving $-1<e^c<1$, but when solving the first inequality, I received an imaginary number $(i\pi)$...

I appreciate any help. Thank you very much

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Well, if $c$ has to be real, then it is always the case that $e^c>0$, so you only have to check for which $c\in\mathbb R$ you have $e^c<1$. The answer is $c<0$.

In any case, if you want $c$ such that $$\sum_{n=0}^\infty e^{nc}=\sum_{n=0}^\infty (e^c)^n=16,$$ think that the aforementioned sum equals $$\frac1{1-e^c},$$ so you need $$\frac1{1-e^c}=16\quad \iff\quad 1=16-16e^c\quad \iff\quad 16e^c=15$$ which is equivalent to $$c=\ln\left(\frac{15}{16}\right).$$

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$$\sum_{n=0}^{+\infty} r^n = \frac{1}{1 - r}$$ if $\vert r \vert < 1$. Let $r = e^c$, so $$\frac{1}{1 - e^c} = 16$$ which gives $$c = \ln \frac{15}{16} $$

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You do not have to worry about the negative side because $e^c$ is a positive number.

All you have to find is the solution to $e^c <1$ which could be done by logarithms.

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