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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

\int e^{c o s (38 t)} sin (38 t) d t

\frac{d}{d x} (x^{2} + 8 x + 7)

y = x, y = 2 x - 15, y = 0

5 \frac{d v}{d t} = 49.05 - 50 v

y = \frac{1}{2} x^{2} + 2 x + 3

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