What is the sum from n=1 to infinity of (4n)/(n^2+5) ?

The sum from n=1 to infinity of (4n)/(n^2+5) is diverges

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sum from n=1 to infinity of (4n)/(n^2+5)

n = 1 \sum \infty \frac{4 n}{n ^{2} + 5}

diverges

Solution steps

n = 1 \sum \infty \frac{4 n}{n ^{2} + 5}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{n}{n ^{2} + 5}

Apply Series Comparison Test:

diverges

= 4 diverges

= diverges

n = 0 \sum \infty (\frac{5}{6})^{n - 1}

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

n = 0 \sum \infty \frac{6}{7 ^{n}}

x = 1 \sum \infty \frac{2 ^{x}}{5 ^{x}}

n = 0 \sum \infty cos (6 n)

What is the sum from n=1 to infinity of (4n)/(n^2+5) ?
The sum from n=1 to infinity of (4n)/(n^2+5) is diverges