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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

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

n = 1 \sum \infty \frac{1}{( n + 3 ) !}

n = 0 \sum \infty \frac{1}{n ln ^{2} ( n )}

n = 1 \sum \infty 70 (0.1)^{n}

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