Telescoping Series Calculator. Enter a summand expression, variable, start value, and end value (or infinity) to evaluate a telescoping series. The calculator identifies the telescoping pattern, cancels consecutive terms, and returns the partial or infinite sum along with convergence information. Also try the calculate Power Series Series Approximation at x.
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. See also our Sigma Notation Calculator.
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)]. You might also find our calculate Fibonacci Number F(n), Golden Ratio Approximation (F(n)/F(n-1)) & Terms Generated — Fibonacci Sequence useful.