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

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

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

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

diverges

Solution steps

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

Expand

\frac{n - 1}{n ^{2}} : \frac{1}{n} - \frac{1}{n ^{2}}

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

Apply Series Limit Comparison Test:

diverges

= diverges

\int 1 + e^{t} d t

y^{^{''}} + 0.1 y^{^{'}} + 0.0004 y = 0

f (x) = \frac{x - 6}{x + 2}, at x = 2

8 x - 3 x

(\frac{1}{2} - \frac{3}{2} \cdot \frac{10}{9} + 4)^{^{'}}

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