Sum of 1/n^4: Does It Converge, and to What Value?

Series convergence, decided and then summed

An infinite series either settles on a number or it does not, and the answer depends on how fast its terms shrink. This series convergence calculator names the test that decides it for four families (geometric, p-series, alternating p-series and telescoping), gives the exact sum when there is one, adds up to a million terms for the partial sum S_N, and tells you how far S_N still is from the limit. Type an accuracy such as 0.001 and it also returns the number of terms that guarantees it.

Quick start. Choose "p-series", type 2 for p and 100 for N. The verdict is converges to π²/6 = 1.644934067, the first 100 terms add up to 1.6349839, and the gap is 0.00995. Add 0.001 as the accuracy and the answer is 1,001 terms.

The four boxes, in the order you meet them

Kind of series. Geometric a + ar + ar² + ..., the p-series Σ 1/n^p (p = 1 is the harmonic series), the alternating p-series Σ (−1)ⁿ⁺¹/n^p, or the telescoping Σ 1/(n(n + k)).
Its parameters. The first term a and the ratio r for a geometric series (fractions such as 1/3 are accepted), the power p for the two p-series, or the whole number k for the telescoping one.
N, the number of terms to add for the partial sum, from 1 to 1,000,000. The table shows S_n at n = 1 to 5, 10, 20, 50, 100 and so on up to N.
Accuracy (optional). A small number such as 0.001. For a convergent series the result adds the smallest N that is certain to bring the error below it.
Read the verdict in the dark box, the three tiles (partial sum, error, terms needed), the reason in plain words, and the table of partial sums.

Terms are always numbered from n = 1. A geometric series that your textbook starts at n = 0 is the same series; just use its first term as a.

Nine series from textbooks and real problems

Example 1: halfway, then half the rest

Situation: walk half a distance, then half of what is left, and so on. Geometric, a = 0.5, r = 0.5, N = 20
Result: converges to 1. After 20 steps the sum is 0.9999990463 and the gap is 9.537 × 10⁻⁷.

An infinite number of steps covers a finite distance. The gap halves with every step, which is what |r| < 1 means.

Example 2: a bouncing ball

Situation: a ball dropped from 2 m rebounds to 60% of each height: 1.2 m, 0.72 m, ... Geometric, a = 1.2, r = 0.6, accuracy 0.01
Result: the rebound heights add up to 3 m, so the total path is 2 + 2 × 3 = 8 m. Twelve bounces bring the sum within 0.01 m.

Each rebound is traveled twice, up and down, which is why the series sum is doubled.

Example 3: a repeating decimal as a fraction

Situation: 0.272727... is 0.27 + 0.0027 + 0.000027 + ... Geometric, a = 0.27, r = 0.01
Result: the sum is 0.27 / 0.99 = 0.2727272727, which is 27/99 = 3/11.

Any repeating block of d digits gives r = 10^−d, so every repeating decimal is a fraction.

Example 4: the Basel problem

Situation: 1 + 1/4 + 1/9 + 1/16 + ... p-series, p = 2, N = 100, accuracy 0.001
Result: converges to π²/6 = 1.644934067. S₁₀₀ = 1.6349839, still 0.00995 short, and a guaranteed error below 0.001 needs 1,001 terms.

The remainder after N terms is close to 1/N here, so each extra digit of accuracy costs ten times as many terms.

Example 5: Apéry's constant

Situation: Σ 1/n³, which has no known closed form. p-series, p = 3, N = 20, accuracy 0.0001
Result: ζ(3) = 1.202056903. Twenty terms give 1.200867842, and 71 terms are enough for an error below 0.0001.

The calculator gets the limit from the Euler-Maclaurin formula, not from adding terms, so the sum is right even when N is small.

Example 6: the harmonic series never stops growing

Situation: 1 + 1/2 + 1/3 + ... p-series, p = 1, N = 100
Result: diverges. S₁₀₀ = 5.187377518, close to the estimate ln 100 + γ + 1/200 = 5.1873859. Passing 10 takes 12,367 terms.

The terms go to 0, yet the sum grows without bound. A term that tends to 0 is necessary for convergence, never sufficient.

Example 7: the alternating harmonic series

Situation: 1 − 1/2 + 1/3 − 1/4 + ... Alternating, p = 1, N = 20, accuracy 0.001
Result: converges conditionally to ln 2 = 0.6931471806. S₂₀ = 0.6687714032, and the Leibniz bound asks for 1,000 terms to be sure of three decimals.

Flip every minus to a plus and you are back at the divergent harmonic series. That is what "conditionally" means.

Example 8: a telescoping series

Situation: 1/(1·2) + 1/(2·3) + 1/(3·4) + ... Telescoping, k = 1, N = 99
Result: converges to 1. S₉₉ = 0.99 exactly, because S_N = 1 − 1/(N + 1). With k = 3 the sum is H₃/3 = 11/18 = 0.6111111111.

Splitting 1/(n(n + k)) into (1/k)(1/n − 1/(n + k)) makes almost every fraction cancel its neighbor k places later.

Example 9: two ways to fail

Situation: 1 − 1 + 1 − 1 + ... (geometric, a = 1, r = −1) and Σ 1/√n (p-series, p = 0.5, N = 10,000).
Result: the first has partial sums that jump between 1 and 0 and never settle; the second creeps upward, reaching 198.5446454 after 10,000 terms and still growing.

A divergent series does not have to run off to infinity. Bounded but restless partial sums diverge too.

Sums of p-series and alternating p-series

Both columns come from the calculator. The alternating sum is always smaller, and the gap between S₁₀ and the limit shows how much faster the alternating series settles.

p	Σ 1/n^p	Error of S₁₀	Σ (−1)ⁿ⁺¹/n^p	Error of S₁₀
0.5	diverges	none	0.6048986434	conditional
1	diverges (harmonic)	none	ln 2 = 0.6931471806	conditional
1.5	2.612375349	0.617	0.7651470246	0.01463
2	π²/6 = 1.644934067	0.09517	π²/12 = 0.8224670334	0.004505
3	1.202056903	0.004525	0.9015426774	0.0004262
4	π⁴/90 = 1.082323234	0.0002867	7π⁴/720 = 0.9470328295	0.00004024
5	1.036927755	0.00002041	0.9721197704	0.000003791
6	π⁶/945 = 1.017343062	0.00000155	0.9855510913	3.564 × 10⁻⁷

Which convergence test fits which series

The calculator applies the first test that settles each of its four families. For a series outside them, this is the usual order to try.

Test	What you check	Typical series
n-th term	if a_n does not tend to 0, the series diverges	Σ n/(n + 1), Σ (−1)ⁿ
Geometric	\|r\| < 1 converges to a/(1 − r)	Σ 3(0.5)ⁿ⁻¹
p-series	p > 1 converges, p ≤ 1 diverges	Σ 1/n², Σ 1/√n
Integral	∫₁^∞ f(x) dx finite, f positive and decreasing	Σ 1/(n ln² n)
Comparison	0 ≤ a_n ≤ b_n with Σ b_n convergent	Σ 1/(n² + 1)
Ratio	lim \|a_n+1/a_n\| < 1 converges, > 1 diverges	Σ 2ⁿ/n!
Root	lim \|a_n\|^1/n < 1 converges	Σ (n/(2n + 1))ⁿ
Alternating (Leibniz)	signs alternate, \|a_n\| decreases to 0	Σ (−1)ⁿ⁺¹/n
Telescoping	terms split into b_n − b_n+k, and b_n tends to 0	Σ 1/(n(n + 1))

How many terms a given accuracy costs

Convergence says nothing about speed, and the four families differ wildly. A geometric series gains a fixed number of digits per term: with r = 0.5 each term halves the gap, so 1 + 1/2 + 1/4 + ... is within a millionth after about 20 terms. A p-series loses the race badly. Its remainder after N terms lies between 1/((p − 1)(N + 1)^p−1) and 1/((p − 1)N^p−1), which for p = 2 means roughly 1/N: three correct decimals need a thousand terms, six need a million.

Alternating series sit in between and come with the simplest guarantee of all. The error after N terms is smaller than the first term you left out, and the true sum always lies between two consecutive partial sums. For the alternating harmonic series that bound is 1/(N + 1), so the 1,000 terms in Example 7 are what three decimals cost. Averaging two consecutive partial sums usually gets far closer than either one, which is the idea behind series acceleration.

The "terms that guarantee" tile uses these bounds, not the actual error, so the true error at that N is usually smaller. For geometric and telescoping series the remainder is known exactly and the count is the smallest N that works.

Where infinite sums turn up outside class

Money. A payment of 100 every year for ever, discounted at 5%, is worth 100/1.05 + 100/1.05² + ..., a geometric series with r = 1/1.05. Its sum is 100/0.05 = 2,000, the textbook value of a perpetuity. The same series with a finite number of terms is an ordinary annuity.

Probability. The chance that a fair coin first shows heads on an odd-numbered toss is 1/2 + 1/8 + 1/32 + ..., geometric with a = 1/2 and r = 1/4, so it is 2/3. Waiting times, expected values of repeated trials and many "who wins first" games reduce to geometric series in the same way.

Physics and computing. A bouncing ball, a signal echoing between two mirrors and the total delay of a buffer that halves each round are geometric. Terms of the form 1/n² set the energy levels of the hydrogen atom, and the harmonic numbers 1 + 1/2 + ... + 1/n explain why the average number of comparisons in quicksort grows like n ln n.

Series questions from calculus class

What is the difference between a sequence and a series?

A sequence is a list of terms a₁, a₂, a₃, ... A series adds them. The series converges when the sequence of partial sums S₁, S₂, S₃, ... has a limit, and that limit is the sum of the series.

If the terms go to zero, does the series converge?

Not necessarily. The harmonic series has terms 1/n that go to zero, yet its partial sums pass every number: 5.187377518 after 100 terms, above 10 after 12,367. Terms going to zero is required for convergence, but it does not prove it.

What does "converges conditionally" mean?

The series converges as written, but the series of absolute values diverges. The alternating harmonic series is the standard case. Such series can be rearranged to add up to any number you like (the Riemann rearrangement theorem), so the order of the terms matters.

Why does the calculator show a sum even when N is small?

The limit comes from a formula (a/(1 − r), the zeta function, the eta function or H_k/k), not from the terms you asked it to add. N only sets the partial sum, so you can see how far S_N is from the limit.

Is 1 − 1 + 1 − 1 + ... equal to 1/2?

Not as an ordinary sum. The partial sums alternate between 1 and 0, so there is no limit and the series diverges. The value 1/2 comes from Cesàro summation, which averages the partial sums; that is a different, weaker notion of a sum.

Can I enter my own formula for a_n?

Not here. A reliable verdict needs a proof, and a proof depends on the form of the terms; adding many terms numerically can make a divergent series look convergent (the harmonic series grows by less than 0.001 per term after term 1000). For a series of your own, compare it with one of the four families using the test table above.

What is the zeta function?

ζ(p) is the sum of 1/n^p for p > 1. Euler found ζ(2) = π²/6 and closed forms for every even p; for odd p such as 3 no closed form is known, and ζ(3) = 1.202056903 was only proved irrational by Apéry.

What about a geometric series with a negative ratio?

With −1 < r < 0 it still converges to a/(1 − r), with partial sums that overshoot and undershoot the limit in turn. At r = −1 they jump between a and 0, and below −1 they swing further out each time. Neither case has a sum.

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What is the sum of 1/n^4, and does the series converge?

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