What is the sum from n=1 to infinity of 4/(7n-1) ?

The sum from n=1 to infinity of 4/(7n-1) is diverges

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sum from n=1 to infinity of 4/(7n-1)

n = 1 \sum \infty \frac{4}{7 n - 1}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{7 n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{7 n - 1}

Apply Series Comparison Test:

diverges

= 4 diverges

= diverges

n = 9 \sum \infty \frac{4}{8 ^{n}}

n = 0 \sum \infty \frac{n ^{2}}{1}

k = 1 \sum \infty 4^{- k}

k = 0 \sum \infty \frac{1}{900 + 3 k}

n = 1 \sum \infty 6^{- 2 n}

What is the sum from n=1 to infinity of 4/(7n-1) ?
The sum from n=1 to infinity of 4/(7n-1) is diverges