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

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

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

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

diverges

Solution steps

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

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

L < 1,

by the Raabe's test

= diverges

\int \frac{x}{x ^{2} + x} d x

\frac{d}{d x} (x^{2} \cdot e^{- x^{2}})

f (x) = - x^{2} - 2 x - 2, at x = - 3

\int \frac{I n ( x )}{x ^{2}} d x

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