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

The sum from n=1 to infinity of (5/2)^n is diverges

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

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

diverges

Solution steps

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

n \to \infty lim (n (\frac{( \frac{5}{2} ) ^{n}}{( \frac{5}{2} ) ^{n + 1}} - 1)) = - \infty

L < 1,

by the Raabe's test

= diverges

\frac{d}{d x} (\frac{( x ^{3} + 8 )}{2 x})^{7}

u^{6}

x \to 0 lim ((\frac{1}{x})^{x})

\int \frac{1}{( 3 x + 2 ) ^{2}} d x

n = 2 \sum \infty \frac{n + 4}{n !}

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