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

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

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

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

converges

Solution steps

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

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} + 4 n + 2}

Apply Series Comparison Test:

converges

= 2 converges

= converges

\frac{d}{d x} (\frac{sin ( 5 x )}{cos ( 6 x )})

\frac{d}{d x} ((x^{2} - 4)^{2} (x - 1))

\frac{( 1 - 5 t )}{2 + t}

\frac{\partial}{\partial x} (\frac{1}{2 π t} e^{- \frac{( x - y ) ^{2}}{2 t}})

x \to 1 - lim (2 x - 1)

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