What is the sum from n=1 to infinity of 2^{n+1} ?

The sum from n=1 to infinity of 2^{n+1} is diverges

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sum from n=1 to infinity of 2^{n+1}

n = 1 \sum \infty 2^{n + 1}

diverges

Solution steps

n = 1 \sum \infty 2^{n + 1}

Apply Series Ratio Test:

diverges

= diverges

n = 0 \sum \infty n \cdot 2^{n}

n = 1 \sum \infty (- π)^{n}

n = 2 \sum \infty n e^{- 4 n}

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

n = 2 \sum \infty (\frac{3}{4})^{3 n}

What is the sum from n=1 to infinity of 2^{n+1} ?
The sum from n=1 to infinity of 2^{n+1} is diverges