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

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

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

n = 2 \sum \infty \frac{- 2}{n + 1}

diverges

Solution steps

n = 2 \sum \infty \frac{- 2}{n + 1}

Apply the constant multiplication rule:

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

= - 2 \cdot n = 2 \sum \infty \frac{1}{n + 1}

Apply Series Limit Comparison Test:

diverges

= - 2 diverges

= diverges

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

n = 0 \sum \infty 8 ln (4 - n)

n = 1 \sum \infty (- \frac{5}{6})^{n - 1}

n = 2 \sum \infty (\frac{3}{8})^{3 n}

n = 0 \sum \infty (1 + \frac{6}{n})^{n}

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