What is the sum of the infinite series ∑ 1/(n(n+1)) from n=1 to ∞?
A1/2 — only the first term survives after cancellation
B1 — the partial sum S_N = 1 - 1/(N+1), which approaches 1
CThe series diverges — partial fractions do not produce convergence
D2 — the first and last terms both survive and each contributes 1
After decomposing 1/(n(n+1)) = 1/n - 1/(n+1), the partial sum collapses to S_N = 1 - 1/(N+1). As N → ∞, 1/(N+1) → 0, leaving the exact sum of 1. Option A is a common error from assuming only one term survives; in this case S_N simplifies cleanly to 1.
Question 2 Multiple Choice
A series has general term aₙ = 1/n - 1/(n+2). After writing out the partial sum S_N, which terms survive the telescoping cancellation?
AOnly 1/1, the very first term
B1/1 and 1/2 from the start, minus 1/(N+1) and 1/(N+2) from the end
CEvery other term — the odd-indexed ones survive
DOnly the final term 1/(N+2)
Writing out S_N = (1/1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + ... reveals that 1/3 is added and subtracted, as is 1/4, 1/5, and so on. Two terms survive at the beginning (1/1 and 1/2) and two at the end (−1/(N+1) and −1/(N+2)). The shift by 2, not 1, means two pairs remain. Always write out the partial sum explicitly rather than guessing which terms survive.
Question 3 True / False
For any telescoping series where aₙ = f(n) − f(n+1), the sum equals f(1) − lim_{N→∞} f(N+1), provided that limit is finite.
TTrue
FFalse
Answer: True
The N-th partial sum is S_N = [f(1)−f(2)] + [f(2)−f(3)] + ... + [f(N)−f(N+1)] = f(1) − f(N+1). The infinite sum is the limit of S_N as N → ∞, which equals f(1) − L whenever f(N+1) → L. This formula is the central result of the telescoping method.
Question 4 True / False
Any series whose terms can be decomposed by partial fractions is a telescoping series.
TTrue
FFalse
Answer: False
Partial fractions is a technique for rewriting terms, but the result is a telescoping series only if the decomposition produces a form like f(n) − f(n+1) so that adjacent terms cancel. For example, 1/((n+1)(n+3)) decomposes into (1/2)(1/(n+1) − 1/(n+3)), which does telescope (with a shift of 2). But not every partial fraction decomposition aligns terms to cancel — it depends entirely on the structure of the shift in the denominator.
Question 5 Short Answer
Why is writing out the partial sum S_N term-by-term essential to evaluating a telescoping series, rather than just recognizing 'it telescopes' and applying the formula directly?
Think about your answer, then reveal below.
Model answer: Writing out S_N explicitly reveals exactly which terms survive at each end of the sum. The number of surviving terms depends on the shift in the formula (a shift of 1 leaves one term at each end; a shift of 2 leaves two). Skipping this step leads to errors in identifying the boundary terms. The written-out sum also confirms the cancellation pattern and catches sign errors in the partial fraction decomposition.
Students who try to apply the formula mechanically without writing out terms often misidentify the surviving endpoint terms, especially when the telescoping involves a shift greater than 1. The physical act of writing the sum and crossing out matching pairs is not a shortcut to skip — it is the derivation itself.