What is the sum from n=1 to infinity of (6/4)^n ?

The sum from n=1 to infinity of (6/4)^n is diverges

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sum from n=1 to infinity of (6/4)^n

n = 1 \sum \infty (\frac{6}{4})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{6}{4})^{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}{4})^{n} - (\frac{6}{4})^{0}

n = 0 \sum \infty (\frac{6}{4})^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int (1 - x) d x

\frac{\partial}{\partial x} (sin (4 x - 7 y))

\frac{( x + 1 ) d y}{d x} = - y + 10

\frac{d}{d x} (x^{2} + 45)

x \to \infty lim (8, x)

What is the sum from n=1 to infinity of (6/4)^n ?
The sum from n=1 to infinity of (6/4)^n is diverges