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

The sum from n=1 to infinity of (2/3)^{n+1} is converges

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sum from n=1 to infinity of (2/3)^{n+1}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

f (x) = sin (x^{2} + 2 x + 3)

y = 2 x^{2} - 1, [0, 1]

y = \frac{x ^{2}}{2 - x ^{2}}

y^{^{''}} + 4 y = x sin (2 x)

\frac{d}{d x} (6 + 4 x - x^{2})

What is the sum from n=1 to infinity of (2/3)^{n+1} ?
The sum from n=1 to infinity of (2/3)^{n+1} is converges