What is the sum from n=1 to infinity of 7/(n^2+4) ?

The sum from n=1 to infinity of 7/(n^2+4) is converges

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sum from n=1 to infinity of 7/(n^2+4)

n = 1 \sum \infty \frac{7}{n ^{2} + 4}

converges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{2} + 4}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 4}

Apply Series Comparison Test:

converges

= 7 converges

= converges

x \to 0 lim (\frac{9 ^{x} - 7 ^{x}}{x})

\int_{1}^{7} \frac{1}{2 x + 2} d x

\frac{d}{d x} (tan^{2} (x^{3}))

x + 1 y = 3, x + 9 = y^{2}

\int \frac{x ^{3} + x ^{2}}{x ^{2} - 9 x + 20} d x

What is the sum from n=1 to infinity of 7/(n^2+4) ?
The sum from n=1 to infinity of 7/(n^2+4) is converges