Math Buddyv1.0

Power series calculator with steps

Type or scan a power series. MathBuddy finds the radius and interval of convergence, step by step.

Loading math input...

What this page covers

  • The radius of convergence RR of a series ∑cn(x−a)n\sum c_n (x-a)^n, found with the ratio test.
  • The interval of convergence, with each endpoint settled separately.
  • Sums of standard power series, such as a geometric series and its derivative, at a given value of x.
  • Not covered here: building a series from a function's derivatives (use Taylor series calculator) and series of plain numbers (use Series convergence calculator).

How to enter the problem

  • Type the series with the sigma key on the math keyboard, then say what you want, such as radius of convergence or interval of convergence.
  • 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 radius of convergence of ∑n=1∞2nxnn\displaystyle \sum_{n=1}^{\infty} \frac{2^n x^n}{n}.

Answer

Verified

The answer is

R=12R = \frac{1}{2}

Explanation

  1. Name the coefficients
    The series has the form ∑cnxn\sum c_n x^n with cn=2nnc_n = \frac{2^n}{n}.
  2. Take the ratio of the coefficients
    ∣cn+1cn∣=2n+1n+1⋅n2n=2nn+1\left|\frac{c_{n+1}}{c_n}\right| = \frac{2^{n+1}}{n+1} \cdot \frac{n}{2^n} = \frac{2n}{n+1}, which approaches L=2L = 2.
  3. Read off the radius
    R=1L=12R = \frac{1}{L} = \frac{1}{2}, so the series converges for ∣x∣<12|x| < \frac{1}{2}.
  4. Note the endpoints
    At x=12x = \frac{1}{2} the series is ∑1n\sum \frac{1}{n}, which diverges; at x=−12x = -\frac{1}{2} it is ∑(−1)nn\sum \frac{(-1)^n}{n}, which converges. The radius is the same either way.

Example 2

Problem

Find the interval of convergence of ∑n=1∞n(x−1)n4n\displaystyle \sum_{n=1}^{\infty} \frac{n (x-1)^n}{4^n}.

Answer

Verified

The answer is

(−3,5)(-3, 5)

Explanation

  1. Apply the ratio test
    ∣an+1an∣=n+1n⋅∣x−1∣4\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{n} \cdot \frac{|x-1|}{4}, which approaches ∣x−1∣4\frac{|x-1|}{4}.
  2. Find the radius
    The series converges when ∣x−1∣4<1\frac{|x-1|}{4} < 1, that is ∣x−1∣<4|x - 1| < 4, so R=4R = 4 around the center 1.
  3. Test the endpoints
    At x=5x = 5 the terms are nn, and at x=−3x = -3 they are (−1)nn(-1)^n n. Neither goes to 0, so both endpoint series diverge.
  4. Write the interval
    The series converges for −3<x<5-3 < x < 5, with both endpoints left out.

Example 3

Problem

Find the sum of ∑n=1∞nxn\displaystyle \sum_{n=1}^{\infty} n x^n at x=13\displaystyle x = \frac{1}{3}.

Answer

Verified

The answer is

34\frac{3}{4}

Explanation

  1. Check that the point is inside the interval
    The ratio test gives R=1R = 1, and ∣13∣<1\left|\frac{1}{3}\right| < 1.
  2. Start from the geometric series
    ∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} for ∣x∣<1|x| < 1.
  3. Differentiate, then multiply by xx
    ∑n=1∞nxn−1=1(1−x)2\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}, so ∑n=1∞nxn=x(1−x)2\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}.
  4. Substitute x=13x = \frac{1}{3}
    1/3(2/3)2=1/34/9=34\frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}

Common mistakes

  • Forgetting the endpoints. The ratio test only gives the open interval ∣x−a∣<R|x - a| < R; each endpoint needs its own test.
  • Turning the ratio upside down. If ∣cn+1cn∣→L\left|\frac{c_{n+1}}{c_n}\right| \to L, the radius is 1L\frac{1}{L}, not LL. For ∑nxn2n\sum \frac{n x^n}{2^n}, L=12L = \frac{1}{2} and R=2R = 2.
  • Dropping the center. For ∑cn(x−1)n\sum c_n (x-1)^n the interval is centered at 1, so ∣x−1∣<4|x - 1| < 4 gives (−3,5)(-3, 5), not (−4,4)(-4, 4).

Checks, assumptions and limits

  • Test one point inside and one outside. At x=1x = 1 the runnable example's terms are n2n\frac{n}{2^n}, which shrink to 0; at x=3x = 3 they are n(32)nn\left(\frac{3}{2}\right)^n, which grow.
  • The radius is never negative. It is a number R≥0R \ge 0 or ∞\infty, and R=0R = 0 means the series converges only at its center.
  • A series may converge at both endpoints, at one, or at neither, so the interval can be open, closed or half-open.
  • MathBuddy can make mistakes. Double check important steps.

Related tasks

Frequently asked questions

The radius RR is how far from the center the series is sure to converge. The interval also says what happens at the two endpoints, so it can be open, closed or half-open.

A Taylor series is a power series built from a function's derivatives at one point. A power series with a positive radius is the Taylor series of the function it adds up to, so the two ideas meet.

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

1 math question to try today without signing in, 3 a day signed in, unlimited with Pro.

Ready to solve your next math problem?

Try one problem free in your browser. Create a free account for 3 math questions a day.

Use the web solverGet the Android appGet the iPhone app

Last updated: · MathBuddy