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}

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

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

Simplify

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

Apply Series Divergence Test:

diverges

= diverges

k = 1 \sum \infty k e^{- 6 k^{2}}

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

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

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

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