What is the sum from n=0 to infinity of 4/(2n+3) ?

The sum from n=0 to infinity of 4/(2n+3) is diverges

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sum from n=0 to infinity of 4/(2n+3)

n = 0 \sum \infty \frac{4}{2 n + 3}

diverges

Solution steps

n = 0 \sum \infty \frac{4}{2 n + 3}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{2 n + 3}

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty 9^{- n}

n = 2 \sum \infty \frac{1}{n ln ( n ^{3} )}

n = 0 \sum \infty \frac{i}{n !}

n = 1 \sum \infty 5^{n} i

n = 1 \sum \infty 5 (- \frac{1}{8})^{5 n}

What is the sum from n=0 to infinity of 4/(2n+3) ?
The sum from n=0 to infinity of 4/(2n+3) is diverges