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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

\frac{1}{n ^{2} + 6 n} : \frac{1}{6 n} - \frac{1}{6 ( n + 6 )}

= 2 \cdot n = 1 \sum \infty \frac{1}{6 n} - \frac{1}{6 ( n + 6 )}

Multiply by the conjugate of

\frac{1}{6 n} - \frac{1}{6 ( n + 6 )} : \frac{\frac{1}{36 n ^{2}} - \frac{1}{36 n ^{2} + 432 n + 1296}}{\frac{1}{6 n} + \frac{1}{6 ( n + 6 )}}

= 2 \cdot n = 1 \sum \infty \frac{\frac{1}{36 n ^{2}} - \frac{1}{36 n ^{2} + 432 n + 1296}}{\frac{1}{6 n} + \frac{1}{6 ( n + 6 )}}

Simplify

\frac{\frac{1}{36 n ^{2}} - \frac{1}{36 n ^{2} + 432 n + 1296}}{\frac{1}{6 n} + \frac{1}{6 ( n + 6 )}} : \frac{1}{n ( n + 6 )}

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

n = 2 \sum \infty \frac{1}{n ^{1.1}}

n = 5 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 5 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

n = 0 \sum \infty (\frac{1}{2 - i})^{n}

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

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