What is the sum from n=1 to infinity of (3^n)/(6^n) ?

The sum from n=1 to infinity of (3^n)/(6^n) is 1

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sum from n=1 to infinity of (3^n)/(6^n)

n = 1 \sum \infty \frac{3 ^{n}}{6 ^{n}}

1

Solution steps

n = 1 \sum \infty \frac{3 ^{n}}{6 ^{n}}

Simplify

\frac{3 ^{n}}{6 ^{n}} : \frac{1}{2 ^{n}}

= 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}

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

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

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

Simplify

= 1

n = 1 \sum \infty 8 (0.1)^{n - 1}

n = 0 \sum \infty 4 + \frac{5}{n ^{3}}

n = 1 \sum \infty \frac{1}{4 n ^{2} - 49}

n = 1 \sum \infty 6 n^{2} e^{- n^{3}}

n = 1 \sum \infty \frac{1}{n ^{\frac{5}{4}}}

What is the sum from n=1 to infinity of (3^n)/(6^n) ?
The sum from n=1 to infinity of (3^n)/(6^n) is 1