What is the sum from n=1 to infinity of 1/(4n^2+1) ?

The sum from n=1 to infinity of 1/(4n^2+1) is converges

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sum from n=1 to infinity of 1/(4n^2+1)

n = 1 \sum \infty \frac{1}{4 n ^{2} + 1}

converges

Solution steps

n = 1 \sum \infty \frac{1}{4 n ^{2} + 1}

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{e ^{(\frac{1}{n})}}{n}

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

k = 1 \sum \infty \frac{3}{k !}

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

n = 1 \sum \infty \frac{n !}{( 3 n ) !}

What is the sum from n=1 to infinity of 1/(4n^2+1) ?
The sum from n=1 to infinity of 1/(4n^2+1) is converges