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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 3 \sum \infty n e^{- n^{2}}

n = 1 \sum \infty sin (n^{n})

n = 3 \sum \infty \frac{2}{3 ln ( n )}

n = 1 \sum \infty \frac{1}{9 n ^{4} + 3}

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

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