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

The sum from n=1 to infinity of-1/(10^n) is -1/9

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

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

- \frac{1}{9}

Solution steps

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

Apply the constant multiplication rule:

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

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

n = 0 \sum \infty \frac{1}{1 0 ^{n}} = \frac{10}{9}

= - (\frac{10}{9} - \frac{1}{1 0 ^{0}})

Simplify

- (\frac{10}{9} - \frac{1}{1 0 ^{0}}) : - \frac{1}{9}

= - \frac{1}{9}

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

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

n = 1 \sum \infty 3 (\frac{1}{4})^{n}

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

n = 1 \sum \infty \frac{4 n}{n ^{2} + 5}

What is the sum from n=1 to infinity of-1/(10^n) ?
The sum from n=1 to infinity of-1/(10^n) is -1/9