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

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

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

n = 1 \sum \infty \frac{7 n}{7 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{7 n}{7 n + 1}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{n}{7 n + 1}

Apply Series Divergence Test:

diverges

= 7 diverges

= diverges

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

n = 8 \sum \infty ln (\frac{n}{4})

k = 1 \sum \infty (cos (18))^{k}

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

n = 2 \sum \infty \frac{n}{( ln ( n ) ) n}

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