We need to compute the series:

$\sum_{t=1}^{\infty} \frac{1}{t(t+1)}$

Step 1: Partial Fraction Decomposition

We first express the general term using partial fractions:

$\frac{1}{t(t+1)} = \frac{A}{t} + \frac{B}{t+1}$

Multiplying both sides by $t(t+1)$ to eliminate denominators:

$1 = A(t+1) + Bt$

Expanding and grouping terms:

$1 = At + A + Bt$

$1 = (A + B)t + A$

By comparing coefficients, we get:

1. $A + B = 0$
2. $A = 1$

Solving for $B$:

$B = -1$

Thus, we rewrite the term:

$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$

Step 2: Series Summation (Telescoping Effect)

Now, we sum from $t=1$ to $\infty$:

$\sum_{t=1}^{\infty} \left( \frac{1}{t} - \frac{1}{t+1} \right)$

Writing out the first few terms:

$\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots$

Most terms cancel in pairs, leaving only:

$1 - \lim_{n \to \infty} \frac{1}{n+1}$

Since $\lim_{n \to \infty} \frac{1}{n+1} = 0$, we obtain:

$\sum_{t=1}^{\infty} \frac{1}{t(t+1)} = 1$

Final Answer:

$\sum_{t=1}^{\infty} \frac{1}{t(t+1)} = 1$

Would you like a further breakdown or a different explanation?

Here are five related questions to explore further:

1. How do you find the sum of a telescoping series in general?
2. What happens if the series starts at a different $t$-value?
3. How does this result compare to the harmonic series $\sum \frac{1}{t}$?
4. Can we generalize this method to similar fractions, e.g., $\frac{1}{t(t+k)}$?
5. How do we approximate the sum of a series when it doesn't telescope?

Tip: When dealing with series of the form $\frac{1}{t(t+k)}$, try partial fraction decomposition to simplify the sum!