What is the sum from n=1 to infinity of (6^n)/(5^n) ?

The sum from n=1 to infinity of (6^n)/(5^n) is diverges

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sum from n=1 to infinity of (6^n)/(5^n)

n = 1 \sum \infty \frac{6 ^{n}}{5 ^{n}}

diverges

Solution steps

n = 1 \sum \infty \frac{6 ^{n}}{5 ^{n}}

Apply exponent rule:

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

= n = 1 \sum \infty (\frac{6}{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{6}{5})^{n} - (\frac{6}{5})^{0}

n = 0 \sum \infty (\frac{6}{5})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d y}{d x} = 2 y - 4 x e^{x}

x^{2} cos (x)

\int \frac{x}{9 x - 8} d x

sin^{3} (2 x)

What is the sum from n=1 to infinity of (6^n)/(5^n) ?
The sum from n=1 to infinity of (6^n)/(5^n) is diverges