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

The sum from n=0 to infinity of n/(n^2) is diverges

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

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

diverges

Solution steps

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

Cancel the common factor:

n

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

Apply Cauchy's Convergence Condition:

diverges

= diverges

f (x) = \frac{2 e ^{x - 1} + ( 2 x - 1 )}{2 x - 1}

\frac{1}{g ( x ^{2} )}

\int sin^{3} (θ) cos^{2} (θ) d θ

\int 6 e^{4 x} cos (8 x) d x

y = \frac{1}{x - 2}, at x = 6

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