Telescoping Series Calculator

A telescoping series is a series where consecutive terms cancel each other when the series is expanded. For example, ∑ [1/(n) − 1/(n+1)] collapses so that only the first and last terms remain, making it straightforward to find the exact sum.

A series is telescoping if it can be written in the form ∑ (a_n − a_{n+1}). This often appears after applying partial fraction decomposition to a rational expression like 1/(n(n+k)). When you expand the first few terms, you'll notice most intermediate terms cancel in pairs.

They allow exact closed-form evaluation of sums that would otherwise require complex techniques. Because most terms cancel, the entire infinite or finite sum reduces to just two values — the starting term and the limiting term — making the computation elegant and exact.

No. Telescoping works only for series that can be expressed as consecutive differences a_n − a_{n+1}. Many series — such as pure geometric or power series — require different convergence tests and summation techniques.

For an infinite telescoping series ∑_{n=1}^{∞} (a_n − a_{n+1}), the sum equals a(n₀) − lim_{n→∞} a(n). If lim a(n) = 0, the series converges and the sum equals just a(n₀), the first surviving term.

Using partial fractions, 1/(n(n+1)) = 1/n − 1/(n+1). The telescoping series collapses to 1 − lim_{n→∞} 1/(n+1) = 1 − 0 = 1. So ∑_{n=1}^{∞} 1/(n(n+1)) = 1.

Common mistakes include choosing the wrong starting index, forgetting to account for the shift k in expressions like 1/(n(n+k)), and incorrectly applying partial fractions. Always verify that the decomposed form truly telescopes by expanding the first few terms manually.

For 1/(n(n+k)), the partial fraction decomposition gives (1/k)[1/n − 1/(n+k)]. A larger k means fewer terms cancel at each step, leaving a larger trailing block of non-cancelled terms in partial sums. The infinite sum becomes (1/k) × [1/n₀ + 1/(n₀+1) + … + 1/(n₀+k−1)].