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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 6 converges

= converges

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

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

n = 0 \sum \infty n^{2} e^{- n^{3}}

n = 1 \sum \infty \frac{5}{n !}

n = 1 \sum \infty \frac{5}{4 ^{n}}

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