What is the sum from n=1 to infinity of 6/(n^2+1) ?

The sum from n=1 to infinity of 6/(n^2+1) is converges

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sum from n=1 to infinity of 6/(n^2+1)

n = 1 \sum \infty \frac{6}{n ^{2} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{6}{n ^{2} + 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 1}

Apply Series Comparison Test:

converges

= 6 converges

= converges

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

n = 1 \sum \infty \frac{19}{1 + 3 ^{n}}

n = 1 \sum \infty \frac{ln ( n )}{7 n}

n = 4 \sum \infty \frac{5}{3 ^{n}}

n = 0 \sum \infty 7 - n^{4}

What is the sum from n=1 to infinity of 6/(n^2+1) ?
The sum from n=1 to infinity of 6/(n^2+1) is converges