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

The sum from i=1 to infinity of 1/(3^i) is 1/2

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sum from i=1 to infinity of 1/(3^i)

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

\frac{1}{2}

Solution steps

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

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

= i = 0 \sum \infty \frac{1}{3 ^{i}} - \frac{1}{3 ^{0}}

i = 0 \sum \infty \frac{1}{3 ^{i}} = \frac{3}{2}

= \frac{3}{2} - \frac{1}{3 ^{0}}

Simplify

\frac{3}{2} - \frac{1}{3 ^{0}} : \frac{1}{2}

= \frac{1}{2}

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

k = 1 \sum \infty \frac{1}{k ^{2}}

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

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

n = 1 \sum \infty (\frac{99}{100})^{n}

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