What is the sum from n=1 to infinity of (9n)/(9n+1) ?

The sum from n=1 to infinity of (9n)/(9n+1) is diverges

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sum from n=1 to infinity of (9n)/(9n+1)

n = 1 \sum \infty \frac{9 n}{9 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{9 n}{9 n + 1}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{n}{9 n + 1}

Apply Series Divergence Test:

diverges

= 9 diverges

= diverges

n = 2 \sum \infty (- 0.8)^{n}

n = 0 \sum \infty 7 (\frac{3}{4})^{n}

n = 0 \sum \infty \frac{5}{n}

n = 0 \sum \infty ∣ n ∣

n = 0 \sum \infty (\frac{7}{5})^{n}

What is the sum from n=1 to infinity of (9n)/(9n+1) ?
The sum from n=1 to infinity of (9n)/(9n+1) is diverges