What is the sum from n=0 to infinity of 1/(n^2+n+1) ?

The sum from n=0 to infinity of 1/(n^2+n+1) is converges

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sum from n=0 to infinity of 1/(n^2+n+1)

n = 0 \sum \infty \frac{1}{n ^{2} + n + 1}

converges

Solution steps

n = 0 \sum \infty \frac{1}{n ^{2} + n + 1}

Apply Series Comparison Test:

converges

= converges

x \to 0 lim (ln (x) + \frac{1}{x})

\int 2 e^{- 2 x} d x

\frac{x d y}{d x} = - y + 2 x^{2} y

\int \frac{5 x - 7 x ^{2} + 4}{( x - 2 ) ^{2} ( x + 1 )} d x

\frac{d}{d x} (e^{- \frac{3 x}{2}})

What is the sum from n=0 to infinity of 1/(n^2+n+1) ?
The sum from n=0 to infinity of 1/(n^2+n+1) is converges