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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 1 \sum \infty - \frac{1}{1 0 ^{n}}

n = 1 \sum \infty 4 (\frac{π}{4})^{n}

n = 1 \sum \infty n! e^{- 2 n}

n = 1 \sum \infty 3 (\frac{1}{4})^{n}

n = 1 \sum \infty \frac{1}{3 n ^{4} + 12 n}

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