What is the sum from n=1 to infinity of 1/3 (e/2)^n ?

The sum from n=1 to infinity of 1/3 (e/2)^n is diverges

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sum from n=1 to infinity of 1/3 (e/2)^n

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 1 \sum \infty (\frac{e}{2})^{n}

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

= \frac{1}{3} (n = 0 \sum \infty (\frac{e}{2})^{n} - (\frac{e}{2})^{0})

n = 0 \sum \infty (\frac{e}{2})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the sum from n=1 to infinity of 1/3 (e/2)^n ?
The sum from n=1 to infinity of 1/3 (e/2)^n is diverges