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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty n! (- e)^{- n}

n = 0 \sum \infty 2 (\frac{2}{e})^{n}

n = 0 \sum \infty 4 (\frac{2}{3})^{n}

n = 1 \sum \infty \frac{n ^{4}}{3 n !}

n = 0 \sum \infty \frac{n ^{5}}{3 ^{n}}

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