What is the sum from n=1 to infinity of (5n)/(n^2+9) ?

The sum from n=1 to infinity of (5n)/(n^2+9) is diverges

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sum from n=1 to infinity of (5n)/(n^2+9)

n = 1 \sum \infty \frac{5 n}{n ^{2} + 9}

diverges

Solution steps

n = 1 \sum \infty \frac{5 n}{n ^{2} + 9}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{n}{n ^{2} + 9}

Apply Series Comparison Test:

diverges

= 5 diverges

= diverges

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

n = 1 \sum \infty n^{2} (\frac{5}{8})^{n}

n = 1 \sum \infty \frac{5 ^{n}}{2 n - 1}

k = 0 \sum \infty (- \frac{5}{6})^{k}

n = 1 \sum \infty (\frac{1}{100})^{n}

What is the sum from n=1 to infinity of (5n)/(n^2+9) ?
The sum from n=1 to infinity of (5n)/(n^2+9) is diverges