Math Buddyv1.0

Sequence convergence calculator with steps

Type or scan a sequence. MathBuddy finds its limit, step by step, and says whether it converges or diverges.

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What this page covers

  • Limits of sequences ana_n as n→∞n \to \infty, given by a formula in nn.
  • Fractions of polynomials, roots, exponentials, powers of n and bounded factors such as sine.
  • Dividing by the highest power, rationalizing, the squeeze theorem, and logarithms for variable exponents.
  • Not covered here: whether the sum of the terms converges (use Series convergence calculator) and sequences given by a recursive rule (use Recursive formula calculator).

How to enter the problem

  • Type the general term as a formula in n, such as (3n+1)/(2n-5), or type lim with n→∞ in front of it.
  • Scan it. Fit the one problem in the frame.
  • Choose a photo. Several problems in one photo are solved as a list.
  • Then tap Solve.

Worked examples

Example 1

Problem

Does an=3n+12n−5\displaystyle a_n = \frac{3n + 1}{2n - 5} converge? Find its limit.

Answer

Verified

The answer is

lim⁡n→∞an=32\lim_{n\to\infty} a_n = \frac{3}{2}

Explanation

  1. Divide by the highest power
    Dividing top and bottom by nn gives an=3+1/n2−5/na_n = \frac{3 + 1/n}{2 - 5/n}.
  2. Let nn grow
    1n→0\frac{1}{n} \to 0 and 5n→0\frac{5}{n} \to 0.
  3. Take the limit
    lim⁡n→∞an=32\lim_{n\to\infty} a_n = \frac{3}{2}. The limit is one finite number, so the sequence converges.

Example 2

Problem

Find the limit of an=sin⁡nn\displaystyle a_n = \frac{\sin n}{n}.

Answer

Verified

The answer is

lim⁡n→∞sin⁡nn=0\lim_{n\to\infty} \frac{\sin n}{n} = 0

Explanation

  1. Bound the terms
    −1≤sin⁡n≤1-1 \le \sin n \le 1, so −1n≤sin⁡nn≤1n-\frac{1}{n} \le \frac{\sin n}{n} \le \frac{1}{n} for every n≥1n \ge 1.
  2. Find the limits of the bounds
    Both −1n-\frac{1}{n} and 1n\frac{1}{n} approach 0.
  3. Apply the squeeze theorem
    The sequence is trapped between two sequences that approach 0, so it approaches 0 as well and converges.

Example 3

Problem

Find the limit of an=n1/n\displaystyle a_n = n^{1/n}.

Answer

Verified

The answer is

lim⁡n→∞n1/n=1\lim_{n\to\infty} n^{1/n} = 1

Explanation

  1. Take logarithms
    ln⁡an=ln⁡nn\ln a_n = \frac{\ln n}{n}
  2. Find the limit of the logarithm
    This is an ∞∞\frac{\infty}{\infty} form. By L'Hôpital's rule, lim⁡n→∞ln⁡nn=lim⁡n→∞1/n1=0\lim_{n\to\infty} \frac{\ln n}{n} = \lim_{n\to\infty} \frac{1/n}{1} = 0.
  3. Undo the logarithm
    an=eln⁡an→e0=1a_n = e^{\ln a_n} \to e^0 = 1, so the sequence converges to 1.

Common mistakes

  • Calling n2+n−n\sqrt{n^2+n} - n zero because n2+n\sqrt{n^2+n} is close to nn. Multiply by n2+n+nn2+n+n\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} first: the limit is 12\frac{1}{2}.
  • Saying (−1)n(-1)^n converges because its values stay between −1-1 and 1. Being bounded is not enough: it jumps between −1-1 and 1 forever, so it diverges.
  • Mixing up the sequence and the series. 1n→0\frac{1}{n} \to 0, so the sequence converges, but the series ∑1n\sum \frac{1}{n} diverges.

Checks, assumptions and limits

  • Check numerically with a large nn. For the runnable example at n=106n = 10^6, 1012+106−106≈0.4999999\sqrt{10^{12} + 10^6} - 10^6 \approx 0.4999999, close to 12\frac{1}{2}.
  • A sequence converges only when its limit is one finite number. A limit of ∞\infty, or values that keep jumping, mean it diverges.
  • Here nn counts 1,2,3,…1, 2, 3, \dots. Changing the first few terms never changes the limit.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

A sequence is a list of terms a1,a2,a3,…a_1, a_2, a_3, \dots. A series is the sum of those terms. A sequence can converge while the series of its terms diverges, as 1n\frac{1}{n} shows.

No. (−1)n(-1)^n is bounded but never settles. A sequence that is bounded and also only increases, or only decreases, does converge, by the monotone convergence theorem.

Yes. Use the camera on the one problem, or choose a photo. Several problems in one photo are solved as a list.

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