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

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

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

n = 2 \sum \infty \frac{( n - 1 )}{n}

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

- x^{4} + 2 x^{3} + 4 x^{2}

f (x) = 3 x^{2} + 2

\int 2^{l n (x)} d x

f (x) = x - 2

\int_{0}^{L} sin^{2} (\frac{π x}{L}) d x

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