What is the sum from n=0 to infinity of 3/((n+1)) ?

The sum from n=0 to infinity of 3/((n+1)) is diverges

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sum from n=0 to infinity of 3/((n+1))

n = 0 \sum \infty \frac{3}{( n + 1 )}

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Limit Comparison Test:

diverges

= 3 diverges

= diverges

\frac{x ^{3}}{2} + \frac{1}{6 x}

\frac{d}{d x} (3 cos (x sin (x)))

n = 1 \sum \infty \frac{n}{n ^{2} + 9}

f (x) = 2 x^{2} - 5 x, at x = 3

\int_{0}^{3} 9 x - x^{3} d x

What is the sum from n=0 to infinity of 3/((n+1)) ?
The sum from n=0 to infinity of 3/((n+1)) is diverges