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

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

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

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

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= 2 diverges

= diverges

sin (x) d x + y d y = 0

\int x^{9} e^{x^{5}} d x

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

\frac{d}{d x} (4 - cos (x))

x \to 2 + lim (4 - x^{2})

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