What is the sum from n=1 to infinity of (3/10)^n ?

The sum from n=1 to infinity of (3/10)^n is 3/7

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sum from n=1 to infinity of (3/10)^n

n = 1 \sum \infty (\frac{3}{10})^{n}

\frac{3}{7}

Solution steps

n = 1 \sum \infty (\frac{3}{10})^{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{3}{10})^{n} - (\frac{3}{10})^{0}

n = 0 \sum \infty (\frac{3}{10})^{n} = \frac{10}{7}

= \frac{10}{7} - (\frac{3}{10})^{0}

Simplify

\frac{10}{7} - (\frac{3}{10})^{0} : \frac{3}{7}

= \frac{3}{7}

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

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

k = 5 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

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

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

What is the sum from n=1 to infinity of (3/10)^n ?
The sum from n=1 to infinity of (3/10)^n is 3/7