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How can I find the sum of the convergent series

Sum from n=1 to infinite of $\frac {e^i}{3^{(i-2)}}$. I know it can be split into Sum of (e.3)^1 + sum of e^i/3^-2. But with this I have e^i which is divergent

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I assume by $n$ you meant $i$. In that case, you have

$$\begin{equation}\begin{aligned} \sum_{i=1}^{\infty} \frac{e^i}{3^{i-2}} &= 3^2\sum_{i=1}^{\infty} \left(\frac{e}{3}\right)^i \\ & = 9\left(\frac{\frac{e}{3}}{1 - \frac{e}{3}}\right) \\ & = \frac{9e}{3 - e} \end{aligned}\end{equation}\tag{1}\label{eq1}$$

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$$\sum_{i=1}^{\infty} \frac{e^i}{3^{i-2}} = 3^2 \sum_{i=1}^{\infty} \left( \frac{e}{3} \right)^i$$ and the last is a geometric series with the ratio $<1$.

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