What is the sum from n=1 to infinity of (-3n)/(2n+1) ?

The sum from n=1 to infinity of (-3n)/(2n+1) is diverges

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sum from n=1 to infinity of (-3n)/(2n+1)

n = 1 \sum \infty \frac{- 3 n}{2 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{- 3 n}{2 n + 1}

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= - 3 diverges

= diverges

n = 2 \sum \infty \frac{1}{n ^{5}}

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

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

n = 1 \sum \infty \frac{1}{ln ( 4 ^{n} )}

n = 1 \sum \infty \frac{3}{n ^{5}}

What is the sum from n=1 to infinity of (-3n)/(2n+1) ?
The sum from n=1 to infinity of (-3n)/(2n+1) is diverges