What is the sum from k=1 to infinity of 30(0.1)^k ?

The sum from k=1 to infinity of 30(0.1)^k is 3.33333…

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sum from k=1 to infinity of 30(0.1)^k

k = 1 \sum \infty 30 (0.1)^{k}

3.33333 \dots

Solution steps

k = 1 \sum \infty 30 \cdot 0. 1^{k}

Apply the constant multiplication rule:

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

= 30 \cdot k = 1 \sum \infty 0. 1^{k}

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

= 30 (k = 0 \sum \infty 0. 1^{k} - 0. 1^{0})

k = 0 \sum \infty 0. 1^{k} = 1.11111 \dots

= 30 (1.11111 \dots - 0. 1^{0})

Simplify

= 3.33333 \dots

n = 1 \sum \infty \frac{8}{n} - \frac{8}{n ^{2}}

n = 1 \sum \infty (\frac{π}{n})^{n} \cdot n!

n = 10 \sum \infty n

n = 0 \sum \infty n^{2} e^{- 3 n}

n = 2 \sum \infty \frac{6}{n ( n + 2 )}

What is the sum from k=1 to infinity of 30(0.1)^k ?
The sum from k=1 to infinity of 30(0.1)^k is 3.33333…