What is the sum from i=1 to infinity of 3e^i ?

The sum from i=1 to infinity of 3e^i is diverges

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sum from i=1 to infinity of 3e^i

i = 1 \sum \infty 3 e^{i}

diverges

Solution steps

i = 1 \sum \infty 3 e^{i}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot i = 1 \sum \infty e^{i}

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

= 3 (i = 0 \sum \infty e^{i} - e^{0})

i = 0 \sum \infty e^{i} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

n = 1 \sum \infty (cos (14))^{n}

n = 1 \sum \infty \frac{- 1 ^{n - 1}}{n}

n = 2 \sum \infty \frac{n}{n ^{2} - 2}

n = 0 \sum \infty (\frac{5}{2})^{n - 1}

What is the sum from i=1 to infinity of 3e^i ?
The sum from i=1 to infinity of 3e^i is diverges