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

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

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

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

n = 1 \sum \infty sin (9)

n = 1 \sum \infty \frac{2}{7 ^{n}}

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

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

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

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