What is the sum from i=1 to infinity of e^{-i} ?

The sum from i=1 to infinity of e^{-i} is e/(e-1)-1

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sum from i=1 to infinity of e^{-i}

i = 1 \sum \infty e^{- i}

\frac{e}{e - 1} - 1

Solution steps

i = 1 \sum \infty e^{- i}

Simplify

e^{- i} : \frac{1}{e ^{i}}

= i = 1 \sum \infty \frac{1}{e ^{i}}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= i = 0 \sum \infty \frac{1}{e ^{i}} - \frac{1}{e ^{0}}

i = 0 \sum \infty \frac{1}{e ^{i}} = \frac{e}{e - 1}

= \frac{e}{e - 1} - \frac{1}{e ^{0}}

Simplify

= \frac{e}{e - 1} - 1

n = 1 \sum \infty \frac{( - 3 + 2 ) ^{n}}{n}

n = 0 \sum \infty \frac{2}{1 + 3 n}

n = 2 \sum \infty \frac{1}{n + 2}

n = 0 \sum \infty 8 + 4 + 2 + 1

n = 1 \sum \infty (- \frac{9}{8})^{n - 1}

What is the sum from i=1 to infinity of e^{-i} ?
The sum from i=1 to infinity of e^{-i} is e/(e-1)-1