What is the sum from n=1 to infinity of n/(n^3+n^2) ?

The sum from n=1 to infinity of n/(n^3+n^2) is 1

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sum from n=1 to infinity of n/(n^3+n^2)

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

1

Solution steps

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

Simplify

\frac{n}{n ^{3} + n ^{2}} : \frac{1}{n ( n + 1 )}

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

Apply Telescoping Series Test:

1

= 1

y^{^{''}} + 2 y^{^{'}} = x^{2} + e^{- 2 x}

f (x) = x, at x = 64.2

x \to 3 lim (6 x^{2} + 2)

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

\int_{1}^{100} \frac{25}{x ^{3}} d x

What is the sum from n=1 to infinity of n/(n^3+n^2) ?
The sum from n=1 to infinity of n/(n^3+n^2) is 1