Find the partial fraction decomposition.
A) Write the original function on the left side of a new equation, and on the right side, write each factor of the denominator in a new fraction. Put a new constant variable in the numerator of each new fraction to represent the unknown numerators—e.g. A, B, C (do not repeat variables using subscript; this can become confusing).
$$\frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}$$B) Multiply both sides by all factors in the original fraction's denominator, cancelling like terms.
$$n(n + 1) \cdot \left( \frac{1}{n(n + 1)} \right) = \left( \frac{A}{n} + \frac{B}{n + 1} \right) n(n + 1)$$D) You should be left with just the products of the unknown numerators and the factors (on the right side) and just the original numerator (on the left side).
$$1 = A \left( n + 1 \right) + B(n)$$E) Now (switching terms on either side for a conventional arrangement), for each unknown quantity remaining on the right side, substitute a value which will remove the unknown by multiplication by zero.
Setting $n = -1$ $$ A\left((-1) + 1\right) + B(-1) = 1 \\ A(0) + B(-1) = 1 \\ -B = 1 \\ B = -1 $$
Setting $n = 0$ $$ A(0 + 1) + B(0) = 1 \\ A(1) = 1 \\ A = 1 $$
F) Now that you've found the unknown values for the partial fractions, all that's left is to rewrite the partial fraction decomposition using the values you found:
$$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$$Now, with your partially decomposed fractions in hand (or in whatever you typically carry decomposed things) ;), you can find the limit of the sequence of partial sums:
A) First, build the sequence by computing the first few terms.
$$S_k = \sum_{n=1}^{k} \frac{1}{n(n + 1)} = \sum_{n=1}^{k} \left( \frac{1}{n} - \frac{1}{n + 1} \right)$$$$S_k= \left( \frac{1}{1} - \frac{1}{1 + 1} \right) + \left( \frac{1}{2} - \frac{1}{2 + 1} \right) + \left( \frac{1}{3} - \frac{1}{3 + 1} \right) + ... + \left( \frac{1}{n} - \frac{1}{n + 1} \right)$$B) Now, a pattern should emerging for you:
$$S_k = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + ... + \left( \frac{1}{n} - \frac{1}{n + 1} \right)$$$$S_k = \left( 1 - \frac{1}{k + 1} \right)$$Aside from the first and the $n\text{th}$ term, all the terms on the inside cancel.
This is why such terms are called telescoping—they sort of slide out when they're expanded, but once you cancel every term you can, you're left with a very short series.
After all those inside terms cancelled, you're left with the simplified sequence of partial sums:
C) Finally, determine whether the series converges and, if it does, find the sum by taking the limit of the sequence of partial sums, $S_k$:
$$\sum_{n=1}^{\infty} \left( \frac{1}{n(n + 1)} \right) = \lim_{n\to \infty} \{S_k\} = \lim_{n\to \infty} \left( 1 - \frac{1}{n + 1} \right)$$$$\sum_{n=1}^{\infty} = \left( 1 - \frac{1}{\infty + 1} \right) = \left( 1 - \frac{1}{\infty} \right) = (1)$$$$\sum_{n=1}^{\infty} \frac{1}{n(n + 1)} = 1$$So, you see—as $n \to \infty$, $S_k \to 1$, and the series converges.
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