Find a closed form for the sum of e^(1 + 2i/n) as i runs from 1 to n.

Step-by-step solution

1. 1

   Split the exponent to separate the constant part. Using $e^{a+b}=e^{a}e^{b}$,

   $e^{1+\frac{2i}{n}}=e\cdot e^{\frac{2i}{n}}\ \Longrightarrow\ \sum_{i=1}^{n}e^{1+\frac{2i}{n}}=e\sum_{i=1}^{n}\left(e^{2/n}\right)^{i}$

   The $e$ has no $i$ in it, so it comes straight out of the sum.

2. 2

   Identify the geometric structure. Writing $r=e^{2/n}$, the terms are $r,r^{2},\dots,r^{n}$: each is the previous one times the fixed ratio $r$. That makes it a finite geometric series with first term $a=r$ and $n$ terms — note $n$ appears both as the number of terms and inside the ratio.

3. 3

   Apply the finite geometric formula. With $a+ar+\cdots+ar^{n-1}=a\dfrac{r^{n}-1}{r-1}$ and $a=r$:

   $\sum_{i=1}^{n}r^{i}=r\cdot\frac{r^{n}-1}{r-1}$

   This needs $r\neq 1$, which holds since $e^{2/n}>1$ for every positive $n$.

4. 4

   Simplify $r^{n}$ — the reason the answer is clean.

   $r^{n}=\left(e^{2/n}\right)^{n}=e^{2}$

   so

   $\sum_{i=1}^{n}e^{\frac{2i}{n}}=e^{2/n}\cdot\frac{e^{2}-1}{e^{2/n}-1}$

   The $n$-dependence collapses into the single factor $e^{2/n}$ and the denominator.

5. 5

   Multiply back by $e$ and interpret the result.

   $\sum_{i=1}^{n}e^{1+\frac{2i}{n}}=e^{1+\frac{2}{n}}\cdot\frac{e^{2}-1}{e^{2/n}-1}$

   This is exactly $\tfrac{n}{2}$ times a right-endpoint Riemann sum for $\int_{1}^{3}e^{x}dx=e^{3}-e$: multiplying by the width $\tfrac2n$ and letting $n\to\infty$ uses $e^{2/n}-1\approx\tfrac2n$ and recovers $e\left(e^{2}-1\right)=e^{3}-e$ ✓.