What is the sum from n=1 to infinity of (2n)/(3n+5) ?

The sum from n=1 to infinity of (2n)/(3n+5) is diverges

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sum from n=1 to infinity of (2n)/(3n+5)

n = 1 \sum \infty \frac{2 n}{3 n + 5}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{3 n + 5}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{3 n + 5}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

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

n = 0 \sum \infty \frac{1}{3 n ^{2}}

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

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

n = 1 \sum \infty \frac{1}{n ^{7} + 2}

What is the sum from n=1 to infinity of (2n)/(3n+5) ?
The sum from n=1 to infinity of (2n)/(3n+5) is diverges