What is the sum from n=1 to infinity of (99)/(100^n) ?

The sum from n=1 to infinity of (99)/(100^n) is 1

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sum from n=1 to infinity of (99)/(100^n)

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

1

Solution steps

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

Apply the constant multiplication rule:

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

= 99 \cdot n = 1 \sum \infty \frac{1}{10 0 ^{n}}

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

= 99 (n = 0 \sum \infty \frac{1}{10 0 ^{n}} - \frac{1}{10 0 ^{0}})

n = 0 \sum \infty \frac{1}{10 0 ^{n}} = \frac{100}{99}

= 99 (\frac{100}{99} - \frac{1}{10 0 ^{0}})

Simplify

99 (\frac{100}{99} - \frac{1}{10 0 ^{0}}) : 1

= 1

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

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

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

n = 1 \sum \infty \frac{n !}{7 ^{n}}

n = 1 \sum \infty \frac{1}{( 3 n - 1 ) ^{2}}

What is the sum from n=1 to infinity of (99)/(100^n) ?
The sum from n=1 to infinity of (99)/(100^n) is 1