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

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

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

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

\frac{\partial}{\partial x} (6 x^{8} y^{6} + 2 x^{7} y^{4})

f (x) = - tan (x) - 5 cot (x)

\frac{d}{d x} (x (x - 1)^{3})

\int - x^{2} (x + 1)^{3} (x - 1) d x

\frac{\partial}{\partial y} (x^{2} + 2 y^{2})

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