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

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

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

n = 0 \sum \infty n (n + 1) - 1

diverges

Solution steps

n = 0 \sum \infty n (n + 1) - 1

Expand

n (n + 1) : n^{2} + n

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

Apply Series Divergence Test:

diverges

= diverges

\int cos (x + π) d x

\frac{e ^{- s}}{s ^{2} ( s + 2 )}

\int 1 + e^{2 x} d x

\frac{d}{d x} (25 x^{2} e^{2 x} - 16)

0, x^{2}, x + 2

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