Series & Convergence Calculator | Partial Sum & Remainder

📖 Tutorial | Series & Convergence

1. Definition

Adding the sequence u₁,u₂,…,u_n,… in order gives the series Σ(n=1..∞) a_n. Let S_N = Σ(k=1..N) a_k be the partial sum. If S_N tends to a finite value S as N→∞, the series converges to S; otherwise it diverges.

2. Symbols

a_n	the n-th term
S_N	partial sum of first N terms
S	sum of the series (if convergent)
R_N = S−S_N	remainder after N terms

3. How it works

Compute S_N by summing the first N terms;
Compute S_{2N}; the difference estimates the remainder;
Check whether a_{N+1} tends to 0;
If the remainder shrinks fast and the term tends to 0, the series likely converges.

4. Steps

Enter a term expression, e.g. 1/(n^2);
Enter N=1000;
Click Calculate and compare S_N with S_{2N};
A tiny remainder indicates good convergence.

5. Example

Example: sum Σ 1/n².
Solution: at N=1000, S_N≈1.6439, S_2N≈1.6444, remainder ≈0.0005; a_{1001}≈9.98e-7→0. The exact sum π²/6≈1.6449 matches.

6. Pitfalls

A partial sum is only approximate; do not conclude convergence from a few terms alone;
If the term does not tend to 0, the series diverges (term test);
Fast numerical convergence is not a proof—use formal tests for rigor.

❓ FAQ | Series & Convergence

What is the difference between a sequence and a series?

A sequence converges if a_n itself tends to a limit; a series converges if its partial sums S_n tend to a finite value.

Why compute S_{2N}?

Comparing S_N with S_{2N} estimates |S−S_N. The closer they are, the smaller the remainder.

Does a_n→0 guarantee convergence?

No. The harmonic series Σ1/n has a_n→0 yet diverges. a_n→0 is necessary, not sufficient.

Can this tool rigorously decide convergence?

It gives numerical evidence. Rigorous decisions require tests like the ratio or integral test.

Is larger N always better?

Larger N reduces the remainder but raises cost and round-off; a few thousand terms usually shows the trend.