What is the sum from n=0 to infinity}n^{'' of-n^' ?

The sum from n=0 to infinity}n^{'' of-n^' is diverges

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sum from n=0 to infinity}n^{'' of-n^'

n = 0 \sum \infty n^{^{''}} - n^{^{'}}

diverges

Solution steps

n = 0 \sum \infty n^{^{''}} - n^{^{'}}

Apply Series Comparison Test:

diverges

= diverges

n = 0 \sum \infty (\frac{99}{100})^{n}

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

n = 1 \sum \infty (\frac{2}{10})^{n}

n = 0 \sum \infty \frac{2}{3 n + 5}

n = 1 \sum \infty \frac{6 n}{e ^{n}}

What is the sum from n=0 to infinity}n^{'' of-n^' ?
The sum from n=0 to infinity}n^{'' of-n^' is diverges