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

The sum from n=1 to infinity of (3^n)/(2^n) is diverges

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sum from n=1 to infinity of (3^n)/(2^n)

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

diverges

Solution steps

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

Apply exponent rule:

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

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

n \to \infty lim (n (\frac{( \frac{3}{2} ) ^{n}}{( \frac{3}{2} ) ^{n + 1}} - 1)) = - \infty

L < 1,

by the Raabe's test

= diverges

\frac{\partial}{\partial x} (2 x^{0.5} y^{0.5})

3 x^{2} + 8 x - 25

\int 4 x^{2} + 3 d x

w (z) = \frac{3 z}{1 + 2 z}

x \to \infty lim (- 8)

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