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

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

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

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

diverges

Solution steps

n = 0 \sum \infty (n + 1) \cdot 1^{n}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{3}{( 4 n + 5 ) ^{3}}

n = 1 \sum \infty 4 sin (\frac{2}{n})

r = 1 \sum \infty r (2 r - 1)

k = 0 \sum \infty (- \frac{1}{6})^{k}

k = 1 \sum \infty \frac{k}{e ^{3 k}}

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