Calculator
Ready. Choose your assumptions, then click evaluate.
Instantly determine whether your setup proves convergence or divergence for a positive-term infinite series or improper integral using the Direct Comparison Test. Then use the in-depth guide below to master every step.
The direct comparison test is one of the fastest and most reliable ways to analyze whether a positive-term infinite series or improper integral converges or diverges. This page gives you both: an interactive calculator for immediate decisions and a full long-form reference so you understand exactly why each conclusion is valid.
The direct comparison test (DCT) is a theorem for nonnegative quantities. In the series setting, it compares two sequences of nonnegative terms, typically aₙ and bₙ. In the improper integral setting, it compares two nonnegative functions, usually f(x) and g(x) on a tail interval like [A,∞).
These two forms are exactly what the calculator checks. If your assumptions match one of these valid forms, you get a definite mathematical conclusion. If your assumptions do not match, the result is marked inconclusive, which is still useful because it tells you to switch tools (often limit comparison, integral test, ratio test, or a stronger inequality).
Use DCT when you can produce a clean inequality between your target expression and a well-known benchmark. Typical benchmarks include p-series terms such as 1/n^p, geometric behavior like r^n with 0<r<1, or p-type improper integrals such as 1/x^p. If your expression includes sums, constants, radicals, or shifted denominators, you can often bound it above or below by one of these familiar forms.
Examples of favorable structure include:
1/(n²+5n) is often less than 1/n².The calculator needs four inputs:
It then applies strict theorem logic:
0 ≤ target ≤ comparison + comparison converges.0 ≤ comparison ≤ target + comparison diverges.If either nonnegativity or eventual validity is missing, DCT is not safely applicable in its standard form, and the calculator warns you accordingly.
Determine whether ∑ 1/(n²+1) converges.
For all n ≥ 1, we have n²+1 ≥ n², so:
The comparison series ∑ 1/n² is a convergent p-series (p=2>1). Therefore, by direct comparison, ∑ 1/(n²+1) converges.
Determine whether ∑ 1/√n diverges.
For n ≥ 1, √n ≤ n, so:
The harmonic series ∑ 1/n diverges, and it is less than or equal to the target terms. Therefore, by direct comparison, ∑ 1/√n diverges.
If you prove 0 ≤ target ≤ comparison and the comparison diverges, DCT does not decide anything. The target may converge or diverge. The calculator returns inconclusive in this case because theorem conditions for a definite conclusion are not met.
Analyze ∫₁^∞ 1/(x²+3) dx.
Since x²+3 ≥ x² for x ≥ 1:
And ∫₁^∞ 1/x² dx converges. Hence the target integral converges by direct comparison.
Analyze ∫₁^∞ 1/(x+1) dx.
For x ≥ 1, we have x+1 ≤ 2x, so:
Because ∫₁^∞ 1/x dx diverges, multiplying by a positive constant does not change divergence, and the target integral diverges by comparison from below.
When solving problems under time pressure, use this short routine:
1/n^p or 1/x^p).This calculator is designed to mirror this exact workflow so your written solutions become both faster and more rigorous.
No. It solves every problem where DCT assumptions are correctly established. Some expressions are easier with limit comparison, ratio, root, alternating series, or integral test.
It means the inequality may fail for early values but holds for all sufficiently large n (or sufficiently large x). Finite starting terms do not affect convergence/divergence of tails.
Positive constant factors do not change convergence class in these contexts. They can make inequality proofs easier.
You may have a theorem mismatch, such as proving target ≤ diverging comparison. That setup does not force a conclusion with DCT.
Use this page as both a direct comparison test calculator and a complete study reference for direct comparison test convergence and divergence decisions in Calculus II and advanced analysis courses.