What is the sum from n=3 to infinity of 2/(n^2+1) ?

The sum from n=3 to infinity of 2/(n^2+1) is converges

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sum from n=3 to infinity of 2/(n^2+1)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

\int \frac{( ln ^{3} ( x ) )}{x} d x

2 y^{^{'}} - y = e^{\frac{t}{3}}

f (x) = 5 ln (x)

\int_{0}^{2} - 2 x^{3} x^{2} + 4 d x

x \to 3 - lim (\frac{∣ x - 3 ∣}{x + 3})

What is the sum from n=3 to infinity of 2/(n^2+1) ?
The sum from n=3 to infinity of 2/(n^2+1) is converges