What is the sum from n=0 to infinity of (5/2)^{n-1} ?

The sum from n=0 to infinity of (5/2)^{n-1} is diverges

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sum from n=0 to infinity of (5/2)^{n-1}

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

diverges

Solution steps

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

Simplify

(\frac{5}{2})^{n - 1} : (\frac{5}{2})^{n} (\frac{5}{2})^{- 1}

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

Apply the constant multiplication rule:

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

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

(\frac{5}{2})^{- 1} = \frac{2}{5}

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

Apply Series Geometric Test:

diverges

= \frac{2}{5} diverges

= diverges

k = 0 \sum \infty (0.4)^{k}

n = 1 \sum \infty n^{3} \frac{1}{n !}

n = 1 \sum \infty \frac{1}{( n - 1 ) !}

n = 2 \sum \infty \frac{9 n}{ln ( n )}

k = 1 \sum \infty [(\frac{k}{2 ^{k}})]

What is the sum from n=0 to infinity of (5/2)^{n-1} ?
The sum from n=0 to infinity of (5/2)^{n-1} is diverges