Calculating Sums of Series

Calculating the sum of a series means finding an exact value for a convergent series, using techniques such as telescoping, partial fractions, geometric series formulas, known power series and differentiation or integration of series. In brief: Σ 1/(n(n + 1)) = Σ (1/n − 1/(n + 1)) telescopes to 1, Σ n xⁿ = x/(1 − x)² for |x| < 1 (obtained by differentiating the geometric series), and Σ 1/n! = e.

Knowing that a series converges does not tell its sum. Exact sums are known only for special series, and finding them is a matter of recognising the pattern and choosing the right technique. This guide collects the main methods with worked examples: finite sums with formulas for powers of integers, telescoping with partial fractions, geometric-type series and known sums from power series, and gives a table of sums to remember.

How do you sum powers of integers?

Three formulas cover many finite sums:

1 + 2 + … + n = n(n + 1)/2
1² + 2² + … + n² = n(n + 1)(2n + 1)/6
1³ + 2³ + … + n³ = [n(n + 1)/2]²

The sum of cubes equals the square of the sum of the integers. Check for n = 4: squares 1 + 4 + 9 + 16 = 30 = 4 · 5 · 9/6 ✓. Cubes 1 + 8 + 27 + 64 = 100 = 10² ✓.

Example 1

Σ_k=1ⁿ (2k + 3)(k − 1) = Σ (2k² + k − 3) = 2 · n(n + 1)(2n + 1)/6 + n(n + 1)/2 − 3n. For n = 10: 2 · 385 + 55 − 30 = 795. Direct check of the first terms: (5)(0) = 0, (7)(1) = 7, (9)(2) = 18, … consistent with the total.

Example 2

1 · 2 + 2 · 3 + … + n(n + 1) = Σ (k² + k) = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)(n + 2)/3. For n = 5: 5 · 6 · 7/3 = 70. Check: 2 + 6 + 12 + 20 + 30 = 70 ✓.

How does telescoping work?

A telescoping series has terms that can be written as a_n = b_n − b_n+1. Then S_N = b₁ − b_N+1, since the middle terms cancel. The partial fraction decomposition often reveals the pattern.

Example 3

Σ_n=1^∞ 1/(n(n + 1)) = Σ (1/n − 1/(n + 1)). S_N = 1 − 1/(N + 1) → 1.

Example 4

Σ_n=1^∞ 1/(n(n + 2)): 1/(n(n + 2)) = (1/2)(1/n − 1/(n + 2)). The terms cancel with a shift of 2: S_N = (1/2)(1 + 1/2 − 1/(N + 1) − 1/(N + 2)) → 3/4.

Example 5

Σ_n=1^∞ 1/(n(n + 1)(n + 2)) = (1/2) Σ [1/(n(n + 1)) − 1/((n + 1)(n + 2))]. S_N = (1/2)(1/2 − 1/((N + 1)(N + 2))) → 1/4.

Example 6: with logarithms

Σ ln(1 + 1/n) = Σ [ln(n + 1) − ln n]. S_N = ln(N + 1), which tends to infinity. The series diverges, although its terms tend to 0.

How do you sum geometric-type series?

For |r| < 1: Σ_n=0^∞ rⁿ = 1/(1 − r). For series with polynomial factors, use the trick of multiplying by r and subtracting, or differentiate the geometric series.

Example 7: Σ n rⁿ

Differentiating Σ rⁿ = 1/(1 − r) term by term gives Σ n rⁿ⁻¹ = 1/(1 − r)², so Σ n rⁿ = r/(1 − r)². For r = 1/2: (1/2)/(1/4) = 2. This confirms Σ n/2ⁿ = 2.

Example 8: the trick of subtracting

S = 1/2 + 2/4 + 3/8 + 4/16 + … Then S/2 = 1/4 + 2/8 + 3/16 + …. Subtract: S/2 = 1/2 + 1/4 + 1/8 + … = 1. So S = 2.

Example 9: Σ (n²) rⁿ

Applying x d/dx to r/(1 − r)² gives Σ n²rⁿ = r(1 + r)/(1 − r)³. For r = 1/2: (1/2)(3/2)/(1/8) = 6.

How do you use known power series?

Some sums can be recognised as values of well-known functions:

e^x = Σ xⁿ/n!, so Σ 1/n! = e and Σ 2ⁿ/n! = e² (sum from n = 0).
sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …
ln(1 + x) = x − x²/2 + x³/3 − … for −1 < x ≤ 1, so 1 − 1/2 + 1/3 − … = ln 2.
arctan x = x − x³/3 + x⁵/5 − …, so 1 − 1/3 + 1/5 − … = π/4.
1/(1 − x) = Σ xⁿ for |x| < 1.

Example 10

Σ_n=0^∞ (−1)ⁿ/(2n)! = cos 1 ≈ 0.5403. Σ_n=1^∞ (−1)ⁿ⁺¹/(n 2ⁿ) = ln(1 + 1/2) = ln 1.5 ≈ 0.4055.

Example 11: the Basel sum

Σ 1/n² = π²/6 (Euler). It follows, for instance, that Σ over odd n of 1/n² = (3/4) · π²/6 = π²/8, since the even terms form Σ 1/(2k)² = (1/4)(π²/6).

Example 12: a mixed series

Σ_n=1^∞ (2ⁿ + 3ⁿ)/6ⁿ = Σ (1/3)ⁿ + Σ (1/2)ⁿ = (1/3)/(2/3) + (1/2)/(1/2) = 1/2 + 1 = 3/2.

Table of sums to remember

Sums of important series
Series	Sum	Method
Σ_n≥0 rⁿ, \|r\| < 1	1/(1 − r)	geometric
Σ_n≥1 n rⁿ	r/(1 − r)²	differentiate
Σ_n≥1 1/(n(n + 1))	1	telescoping
Σ_n≥1 1/(n(n + 2))	3/4	telescoping
Σ_n≥0 1/n!	e	power series of e^x
Σ_n≥1 1/n²	π²/6	Euler
Σ_n≥1 (−1)ⁿ⁺¹/n	ln 2	power series of ln(1 + x)
Σ_n≥0 (−1)ⁿ/(2n + 1)	π/4	power series of arctan

Where do these sums come from?

Archimedes summed geometric series. Madhava of Sangamagrama, in the fourteenth century, and later Gregory and Leibniz found the series for the arctangent and π/4. Euler solved the Basel problem, the sum of the reciprocals of the squares, in 1735, a problem that had resisted Jacob Bernoulli and others. Telescoping sums were used by Euler and the Bernoullis in the eighteenth century.

Where are they used today?

Exact sums appear in probability (expected values: for a geometric distribution with success probability p, the expected number of trials is Σ n p (1 − p)ⁿ⁻¹ = 1/p), in physics and in the analysis of algorithms. The finite sums of powers give the number of operations in nested loops.

Which mistakes are common?

Cancelling the wrong terms in telescoping

Write out the first few and last few terms of the partial sum. Terms that remain are at the ends.

Ignoring convergence before summing

Formulas such as the geometric sum hold only inside their region of convergence.

Mixing up the starting index

Σ_n=0 rⁿ = 1/(1 − r), but Σ_n=1 rⁿ = r/(1 − r). Check the first term.

Differentiating without care

Differentiating a power series term by term is valid inside its radius of convergence, but the index may shift.