What is the sum from n=2 to infinity of 1/(5^n) ?

The sum from n=2 to infinity of 1/(5^n) is 1/20

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sum from n=2 to infinity of 1/(5^n)

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

\frac{1}{20}

Solution steps

n = 2 \sum \infty \frac{1}{5 ^{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}{5 ^{n}} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}

n = 0 \sum \infty \frac{1}{5 ^{n}} = \frac{5}{4}

= \frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}}

Simplify

\frac{5}{4} - \frac{1}{5 ^{0}} - \frac{1}{5 ^{1}} : \frac{1}{20}

= \frac{1}{20}

x \to 0 lim (\frac{sin ( x )}{8 x})

f (x) = \frac{sin ( x ) + cos ( x )}{e ^{x}}

\int 3 x^{2} - 8 x + C d x

h (x) = (x tan (x))^{7}

\int x sin (a x) d x

What is the sum from n=2 to infinity of 1/(5^n) ?
The sum from n=2 to infinity of 1/(5^n) is 1/20