What is the sum from n=70 to infinity of 3/(n^2+2n) ?

The sum from n=70 to infinity of 3/(n^2+2n) is converges

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sum from n=70 to infinity of 3/(n^2+2n)

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

converges

Solution steps

n = 70 \sum \infty \frac{3}{n ^{2} + 2 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 70 \sum \infty \frac{1}{n ^{2} + 2 n}

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

= 3 \cdot n = 70 \sum \infty \frac{n ^{2} - 2 n}{( n ^{2} ) ^{2} - ( 2 n ) ^{2}}

Simplify

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

= 3 \cdot n = 70 \sum \infty \frac{1}{n ( n + 2 )}

Apply Series Comparison Test:

converges

= 3 converges

= converges

What is the sum from n=70 to infinity of 3/(n^2+2n) ?
The sum from n=70 to infinity of 3/(n^2+2n) is converges