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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

m = 1 \sum \infty \frac{5}{m ln ( 3 m )}

n = 1 \sum \infty \frac{5}{4 ^{n} + 1}

n = 0 \sum \infty \frac{1}{6 ^{n} - 5}

n = 3 \sum \infty \frac{1}{n ( n - 2 )}

n = 5 \sum \infty \frac{1}{n ^{2} ln ( n )}

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