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

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

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

n = 3 \sum \infty \frac{6}{( n ^{2} - 1 )}

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 6 converges

= converges

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

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

n = 1 \sum \infty \frac{8}{n ^{7} + 9}

n = 0 \sum \infty 8 (\frac{7}{8})^{n}

n = 1 \sum \infty 5 n sin (\frac{4}{n})

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