What is the sum from n=4 to infinity of 4(6-2n) ?

The sum from n=4 to infinity of 4(6-2n) is diverges

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sum from n=4 to infinity of 4(6-2n)

n = 4 \sum \infty 4 (6 - 2 n)

diverges

Solution steps

n = 4 \sum \infty 4 (6 - 2 n)

Apply the constant multiplication rule:

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

= 4 \cdot n = 4 \sum \infty 6 - 2 n

Apply Series Divergence Test:

diverges

= 4 diverges

= diverges

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

n = 0 \sum \infty \frac{9}{1 0 ^{n}}

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

n = 1 \sum \infty (\frac{11}{5})^{n}

k = 1 \sum \infty \frac{1}{1 1 ^{k}}

What is the sum from n=4 to infinity of 4(6-2n) ?
The sum from n=4 to infinity of 4(6-2n) is diverges