What is the sum from n=2 to infinity of 2/(7^n) ?

The sum from n=2 to infinity of 2/(7^n) is 1/21

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sum from n=2 to infinity of 2/(7^n)

n = 2 \sum \infty \frac{2}{7 ^{n}}

\frac{1}{21}

Solution steps

n = 2 \sum \infty \frac{2}{7 ^{n}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 2 \sum \infty \frac{1}{7 ^{n}}

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

= 2 (n = 0 \sum \infty \frac{1}{7 ^{n}} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

n = 0 \sum \infty \frac{1}{7 ^{n}} = \frac{7}{6}

= 2 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}})

Simplify

2 (\frac{7}{6} - \frac{1}{7 ^{0}} - \frac{1}{7 ^{1}}) : \frac{1}{21}

= \frac{1}{21}

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

n = 0 \sum \infty \frac{n}{ln ( n ^{2} )}

n = 1 \sum \infty n^{- s i n (1)}

n = 1 \sum \infty cos^{n} (5)

k = 1 \sum \infty (\frac{2}{5})^{k}

What is the sum from n=2 to infinity of 2/(7^n) ?
The sum from n=2 to infinity of 2/(7^n) is 1/21