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

The sum from n=1 to infinity of (1/2)^{n-2} is 4

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

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

4

Solution steps

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

Simplify

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

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

Apply the constant multiplication rule:

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

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

(\frac{1}{2})^{- 2} = 4

= 4 \cdot n = 1 \sum \infty (\frac{1}{2})^{n}

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

= 4 (n = 0 \sum \infty (\frac{1}{2})^{n} - (\frac{1}{2})^{0})

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

= 4 (2 - (\frac{1}{2})^{0})

Simplify

= 4

k = 1 \sum \infty 6 (\frac{1}{5})^{k - 1}

n = 3 \sum \infty \frac{7}{3 ln ( n )}

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

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

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

What is the sum from n=1 to infinity of (1/2)^{n-2} ?
The sum from n=1 to infinity of (1/2)^{n-2} is 4