What is the sum from n=1 to infinity of 3^n ?

The sum from n=1 to infinity of 3^n is diverges

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sum from n=1 to infinity of 3^n

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

diverges

Solution steps

n = 1 \sum \infty 3^{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 3^{n} - 3^{0}

n = 0 \sum \infty 3^{n} =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int \frac{1 + 2 x}{x} d x

n \to \infty lim (3)

\frac{\partial}{\partial x} (cos (y e^{x y}) sin (x))

\int \frac{( a ^{2} - x ^{2} ) ^{\frac{3}{2}}}{x ^{4}} d x

x \to 0 lim (\frac{sin ( 16 x )}{3 x})

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