What is the sum from k=2 to infinity of 3e^{-2k} ?

The sum from k=2 to infinity of 3e^{-2k} is converges

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sum from k=2 to infinity of 3e^{-2k}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 3 \cdot k = 2 \sum \infty e^{- 2 k}

Apply Series Ratio Test:

converges

= 3 converges

= converges

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

n = 1 \sum \infty 60 (0.1)^{n}

n = 0 \sum \infty (\frac{5}{9})^{n}

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

n = 0 \sum \infty 3 + (- 1)^{n}

What is the sum from k=2 to infinity of 3e^{-2k} ?
The sum from k=2 to infinity of 3e^{-2k} is converges