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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty \frac{1}{5 + 6 n}

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

n = 1 \sum \infty 35 (- 1)^{n}

k = 1 \sum \infty (cos (9))^{k}

n = 1 \sum \infty (\frac{6}{16})^{n}

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