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Comparison Test Calculator

Compare with a known series using the limit form to judge a positive-term series.

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L = lim(a_n/b_n) ∈ (0,∞) ⇒ both series converge or both diverge
e.g. 1/(n^2+1)
e.g. 1/(n^2)
1=convergent 0=divergent

📖 Tutorial | Comparison Test

1. Theorem
For positive-term series Σ a_n and Σ b_n, if L = lim(a_n/b_n) exists and 0, the two behave alike. If L=0 and Σ b_n converges, then Σ a_n converges; if L=∞ and Σ b_n diverges, then Σ a_n diverges.
2. Symbols
a_nseries under test
b_ncomparison series
L = a_n/b_nlimit of the ratio
known status of b_nthe reference
3. How it works
  • Approach the limit with large n=100000;
  • Compute a_n/b_n;
  • A positive constant L means same behavior;
  • L=0 favors convergence, L=∞ favors divergence.
4. Steps
  1. Enter a_n, e.g. 1/(n^2+1);
  2. Enter b_n, e.g. 1/(n^2);
  3. Set reference status 1=convergent, 0=divergent;
  4. Click Calculate.
5. Example
Example: Σ 1/(n²+1).
Solution: take b_n=1/n² (convergent). For large n, a_n/b_n = n²/(n²+1)→1>0, so the series converges like b_n.
6. Pitfalls
Both series must have positive terms;
b_n must be a series with known behavior, usually a p-series;
A positive constant L gives the strongest conclusion.

❓ FAQ | Comparison Test

How to choose b_n?
Usually a p-series Σ1/n^p or geometric series matching the leading order of a_n.
What does a positive constant L mean?
The two series are asymptotically the same order, so they share convergence.
How to read L=0?
If the reference converges, a_n converges too; if it diverges, no conclusion.
Is this rigorous?
It approximates the limit for quick judgment; a proof needs an ε-N argument.
Can I use it for non-positive series?
Not directly; test absolute value with comparison first.