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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 5 converges

= converges

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

k = 0 \sum \infty \frac{1}{100 + k}

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

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

n = 0 \sum \infty \frac{14}{14 + n}

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