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

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

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

n = 1 \sum \infty 1 - \frac{1}{n + 1}

diverges

Solution steps

n = 1 \sum \infty 1 - \frac{1}{n + 1}

Apply Series Divergence Test:

diverges

= diverges

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

n = 2 \sum \infty 6 e^{- 3 n}

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

n = 1 \sum \infty \frac{n !}{10 9 ^{n}}

n = 2 \sum \infty \frac{1}{n ^{2} + 4}

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