What is the sum from n=1 to infinity of (3n)/(3n+1) ?

The sum from n=1 to infinity of (3n)/(3n+1) is diverges

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sum from n=1 to infinity of (3n)/(3n+1)

n = 1 \sum \infty \frac{3 n}{3 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{3 n}{3 n + 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{n}{3 n + 1}

Apply Series Divergence Test:

diverges

= 3 diverges

= diverges

n = 0 \sum \infty 7^{n}

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

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

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

n = 0 \sum \infty (\frac{1}{n})^{n}

What is the sum from n=1 to infinity of (3n)/(3n+1) ?
The sum from n=1 to infinity of (3n)/(3n+1) is diverges