What is the sum from n=2 to infinity of 5(-1/3)^n ?

The sum from n=2 to infinity of 5(-1/3)^n is converges

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sum from n=2 to infinity of 5(-1/3)^n

n = 2 \sum \infty 5 (- \frac{1}{3})^{n}

converges

Solution steps

n = 2 \sum \infty 5 (- \frac{1}{3})^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 5 \cdot n = 2 \sum \infty (- \frac{1}{3})^{n}

Apply Series Ratio Test:

converges

= 5 converges

= converges

\int \frac{5 x ^{2}}{2} d x

y = \frac{3 ln ( 4 x )}{5 - 2 x}

f (x) = - 3 x^{3} - 2 x, (- 1, 5)

x^{2} y^{^{'}} + y = 0

n lo g_{10} (n)

What is the sum from n=2 to infinity of 5(-1/3)^n ?
The sum from n=2 to infinity of 5(-1/3)^n is converges