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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 5 converges

= converges

k = 2 \sum \infty 9 e^{- 2 k}

k = 2 \sum \infty (\frac{1}{3})^{3 k}

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

k = 2 \sum \infty (- 0.9)^{k}

k = 0 \sum \infty (\frac{5}{8})^{k}

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