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

The sum from n=1 to infinity of 1/(3^{n-3)} is converges

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y = \frac{2 x + 1}{x + 2}, (1, 1)

\frac{\partial}{\partial y} (cos (2 x y))

g (x) = 9 - 2

\int_{- 1}^{1} \frac{5 x}{( x + 3 ) ^{2}} d x

\int \frac{1}{1 + 4 y ^{2}} d y

What is the sum from n=1 to infinity of 1/(3^{n-3)} ?
The sum from n=1 to infinity of 1/(3^{n-3)} is converges