Math Buddyv1.0

Series convergence calculator with steps

Type or scan a series. MathBuddy picks a convergence test, works it through, and finds the sum when possible.

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What this page covers

  • Infinite series written in sigma notation, starting at any index.
  • The divergence, geometric, p-series, ratio, root, comparison and alternating series tests, with the reason each one applies.
  • The sum of the series when it has a closed form, such as geometric and telescoping series.
  • Not covered here: the limit of a sequence on its own (use Sequence convergence calculator) and the interval of convergence of a power series (use Power series calculator).

How to enter the problem

  • Type it with the sigma key on the math keyboard, or as sum from n = 1 to infinity of the term.
  • 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

Find the limit of the terms of ∑n=1∞n2n+1\displaystyle \sum_{n=1}^{\infty} \frac{n}{2n+1} and decide whether the series converges.

Answer

Verified

The answer is

lim⁡n→∞an=12\lim_{n\to\infty} a_n = \frac{1}{2}

Explanation

  1. Look at the terms
    The terms are an=n2n+1a_n = \frac{n}{2n+1}. If they do not approach 0, the series cannot converge.
  2. Divide by the highest power
    Dividing top and bottom by nn gives an=12+1/na_n = \frac{1}{2 + 1/n}, and 1n→0\frac{1}{n} \to 0 as n→∞n \to \infty.
  3. Take the limit
    lim⁡n→∞an=12\lim_{n\to\infty} a_n = \frac{1}{2}
  4. Apply the divergence test
    The terms approach 12\frac{1}{2}, not 0, so the partial sums keep growing by about 12\frac{1}{2} each time. The series diverges.

Example 2

Problem

Show that ∑n=0∞2n+3n6n\displaystyle \sum_{n=0}^{\infty} \frac{2^n + 3^n}{6^n} converges and find its sum.

Answer

Verified

The answer is

∑n=0∞2n+3n6n=72\sum_{n=0}^{\infty} \frac{2^n + 3^n}{6^n} = \frac{7}{2}

Explanation

  1. Split the term
    2n+3n6n=(13)n+(12)n\frac{2^n + 3^n}{6^n} = \left(\frac{1}{3}\right)^n + \left(\frac{1}{2}\right)^n, so the series is the sum of two geometric series.
  2. Check each ratio
    The ratios are 13\frac{1}{3} and 12\frac{1}{2}. Both are less than 1 in size, so both geometric series converge, and so does their sum.
  3. Sum each geometric series
    Starting at n=0n = 0, ∑rn=11−r\sum r^n = \frac{1}{1-r}. This gives 11−1/3=32\frac{1}{1 - 1/3} = \frac{3}{2} and 11−1/2=2\frac{1}{1 - 1/2} = 2.
  4. Add the two sums
    32+2=72\frac{3}{2} + 2 = \frac{7}{2}

Example 3

Problem

Use the ratio test to show that ∑n=1∞n2n\displaystyle \sum_{n=1}^{\infty} \frac{n}{2^n} converges, then find its sum.

Answer

Verified

The answer is

∑n=1∞n2n=2\sum_{n=1}^{\infty} \frac{n}{2^n} = 2

Explanation

  1. Set up the ratio
    an+1an=n+12n+1⋅2nn=n+12n\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}
  2. Take the limit of the ratio
    lim⁡n→∞n+12n=12\lim_{n\to\infty} \frac{n+1}{2n} = \frac{1}{2}. Since 12<1\frac{1}{2} < 1, the ratio test says the series converges.
  3. Start from a geometric series
    For ∣x∣<1|x| < 1, ∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}. Differentiating and then multiplying by xx gives ∑n=1∞nxn=x(1−x)2\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}.
  4. Substitute x=12x = \frac{1}{2}
    1/2(1/2)2=1/21/4=2\frac{1/2}{(1/2)^2} = \frac{1/2}{1/4} = 2

Common mistakes

  • Taking terms that go to 0 as proof of convergence. The terms of ∑1n\sum \frac{1}{n} go to 0, yet that series diverges. The divergence test can only show divergence.
  • Reading a ratio limit of exactly 1 as a verdict. For ∑1n\sum \frac{1}{n} and ∑1n2\sum \frac{1}{n^2} the ratio limit is 1 in both cases, but the first diverges and the second converges, so another test is needed.
  • Starting a geometric sum at the wrong index: ∑n=1∞(12)n=1\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n = 1, not 2. The value 2 is the sum from n=0n = 0.
  • Confusing the limit of the terms with the sum. In ∑n2n\sum \frac{n}{2^n} the terms go to 0, but the sum is 2.

Checks, assumptions and limits

  • Check a sum with partial sums. For the runnable example, 11⋅2+12⋅3+⋯+1N(N+1)=1−1N+1\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \dots + \frac{1}{N(N+1)} = 1 - \frac{1}{N+1}, which is 99100\frac{99}{100} at N=99N = 99 and approaches 1.
  • Each test has conditions. The alternating series test needs terms that shrink to 0, and the comparison tests need positive terms.
  • Dropping or changing finitely many terms changes the sum but never whether the series converges.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

A sequence converges when its terms approach one number. A series converges when its partial sums, the running totals of the terms, approach one number. The terms of a convergent series must go to 0, but terms going to 0 is not enough on its own.

Start with the divergence test: if the terms do not go to 0, the series diverges. Then match the form. Geometric series and p-series have direct rules, factorials and powers like 2n2^n suit the ratio test, and fractions of polynomials suit a comparison with a p-series.

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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