What is the sum from n=101 to infinity of-1/(n(n-1)) ?

The sum from n=101 to infinity of-1/(n(n-1)) is -1/100

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sum from n=101 to infinity of-1/(n(n-1))

n = 101 \sum \infty - \frac{1}{n ( n - 1 )}

- \frac{1}{100}

Solution steps

n = 101 \sum \infty - \frac{1}{n ( n - 1 )}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 101 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

\frac{1}{100}

= - \frac{1}{100}

x \to 0 - lim (- e^{\frac{1}{x}})

f (x) = 5 x^{2} + 8, (- 3, 53)

\frac{d}{d x} ((a + b - 5) x^{2} + (b + c - 7) x + (a + c))

f (x) = x + \frac{3}{x}, at x = 3

(- 5.6) (- 2.5)

What is the sum from n=101 to infinity of-1/(n(n-1)) ?
The sum from n=101 to infinity of-1/(n(n-1)) is -1/100