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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Limit Comparison Test:

diverges

= 2 diverges

= diverges

y = e^{- 3 x}, x \geq 0

\frac{3}{4 x ^{5}}

x \to \infty lim (1^{2 x})

x \to 0 + lim (\frac{7}{1 + e ^{\frac{1}{x}}})

n = 1 \sum \infty n (n + 1) x^{n}

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