What is the sum from n=1 to infinity of (2n)/(3n+6) ?

The sum from n=1 to infinity of (2n)/(3n+6) is diverges

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sum from n=1 to infinity of (2n)/(3n+6)

n = 1 \sum \infty \frac{2 n}{3 n + 6}

diverges

Solution steps

n = 1 \sum \infty \frac{2 n}{3 n + 6}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n}{3 n + 6}

Apply Series Divergence Test:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty \frac{n}{6 ^{n}}

n = 2 \sum \infty \frac{4}{n ( n - 1 )}

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

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

n = 1 \sum \infty \frac{n !}{6 ^{n}}

What is the sum from n=1 to infinity of (2n)/(3n+6) ?
The sum from n=1 to infinity of (2n)/(3n+6) is diverges