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

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

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

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

diverges

Solution steps

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

Simplify

3 - \frac{2}{3 ^{n}} : \frac{3 ^{1 + n} - 2}{3 ^{n}}

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

Apply Series Divergence Test:

diverges

= diverges

\int_{b}^{2 b} x^{7} d x

h \to + 0 lim (h)

\int 2 x y^{2} - 3 d x

\int \frac{1}{x ^{2} x ^{2} - 8} d x

\frac{\partial}{\partial y} (ln (x^{2} + 3 x y))

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