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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Limit Comparison Test:

diverges

= - diverges

= diverges

\int e^{3 x} (sin (3 x) + cos (3 x)) d x

\int (x^{2} - \frac{3}{x}) d x

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

y = \frac{( 3 x - 6 )}{( x ^{2} + 4 )}

x^{2} + y^{3} = 12

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