What is the sum from n=0 to infinity of (-1+4/n)^4 ?

The sum from n=0 to infinity of (-1+4/n)^4 is diverges

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sum from n=0 to infinity of (-1+4/n)^4

n = 0 \sum \infty (- 1 + \frac{4}{n})^{4}

diverges

Solution steps

n = 1 \sum \infty (- 1 + \frac{4}{n})^{4}

Apply Series Divergence Test:

diverges

= diverges

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

n = 1 \sum \infty (\frac{2}{9})^{n}

n = 1 \sum \infty n e^{- 16 n}

n = 1 \sum \infty (- 0.005)^{n}

n = 1 \sum \infty \frac{2}{n ln ( 9 n )}

What is the sum from n=0 to infinity of (-1+4/n)^4 ?
The sum from n=0 to infinity of (-1+4/n)^4 is diverges