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

The sum from x=1 to infinity of (2^x)/(5^x) is 2/3

Popular >

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

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

\frac{2}{3}

Solution steps

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{2 ^{n}}{5 ^{n}} = (\frac{2}{5})^{n}

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

n = 0 \sum \infty (\frac{2}{5})^{n} = \frac{5}{3}

= \frac{5}{3} - (\frac{2}{5})^{0}

Simplify

\frac{5}{3} - (\frac{2}{5})^{0} : \frac{2}{3}

= \frac{2}{3}

n = 0 \sum \infty cos (6 n)

n = 0 \sum \infty (\frac{1}{23})^{n}

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

n = 1 \sum \infty 3 (\frac{2}{5})^{n}

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

What is the sum from x=1 to infinity of (2^x)/(5^x) ?
The sum from x=1 to infinity of (2^x)/(5^x) is 2/3