What is the sum from n=1 to infinity of (6n)/(6n+1) ?

The sum from n=1 to infinity of (6n)/(6n+1) is diverges

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sum from n=1 to infinity of (6n)/(6n+1)

n = 1 \sum \infty \frac{6 n}{6 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{6 n}{6 n + 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{n}{6 n + 1}

Apply Series Divergence Test:

diverges

= 6 diverges

= diverges

n = 1 \sum \infty \frac{1}{n} - \frac{2}{3 n - 1}

k = 2 \sum \infty (- 0.95)^{k}

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

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

n = 1 \sum \infty n - 1

What is the sum from n=1 to infinity of (6n)/(6n+1) ?
The sum from n=1 to infinity of (6n)/(6n+1) is diverges