Near the very end of this course, some differential equations refuse to hand back a solution in terms of familiar functions at all — and the way around that is to build the solution directly as a power series, term by term. Everything in this section is the machinery that method leans on.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
For \displaystyle\sum_{n=0}^\infty a_n(x-1)^n with a_0=5, what is the value of the series at x=1?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Only the constant term a_0 survives at the center; every other term carries a factor of (1-1)^n=0.
Correct answer
Also accepted:
Why
From Calculus II — Integral Calculus & Infinite Series
True or false — every power series converges at its own center x=c, regardless of the coefficients.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Every term past the first carries a factor of (c-c)^n=0, no matter what the coefficients are — nothing can make that term nonzero.
No boundedness condition is needed — the (x-c)^n=0 factor for n\ge1 holds unconditionally at x=c.
This has nothing to do with whether the series happens to be geometric — it's true for every power series.
Why
At x=c, all terms past n=0 vanish identically, so convergence at the center is automatic and universal.
Plug x=c into a_n(x-c)^n and see what happens to every term past n=0.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Find the radius of convergence of \displaystyle\sum_{n=0}^{\infty}n!\,x^n.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\left|\frac{a_{n+1}}{a_n}\right|=(n+1)|x|\to\infty for any x\neq0, so the series diverges everywhere except x=0 itself — R=0.
Correct answer
Also accepted:
Why
Apply the Ratio Test — what happens to \frac{(n+1)!}{n!} as n\to\infty?
From Calculus II — Integral Calculus & Infinite Series
A power series has radius of convergence R=\infty. What does this mean?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That describes R=0, the opposite extreme.
R=\infty means BOTH directions from the center extend without bound, not just one side.
The Ratio Test is exactly how R=\infty typically gets discovered — the limit L(x) comes out to 0 for every x.
Why
R=\infty means the interval of convergence is all of (-\infty,\infty) — the series converges no matter what real number is substituted for x.
Think about what "the series converges no matter what real number is substituted" would mean for the interval of convergence.
A power series is an infinitely long polynomial. The radius of convergence is how far from its center that infinite sum still adds up to a real, finite number instead of spiraling off to infinity.
A power series \displaystyle\sum_{n=0}^\infty a_n(x-x_0)^n converges for x within some radius R of x_0 and diverges outside it — the ratio test, applied to the series' own terms, is the usual way to find R. Later, a series solution to a differential equation is built by substituting y=\sum a_n x^n directly into the equation and solving for the coefficients a_n one at a time from a recurrence relation. That substitution is only legitimate on the interval where the resulting series actually converges — which is exactly why finding R isn't optional bookkeeping, it's the boundary of where the constructed "solution" means anything at all.
Where does this series actually converge?
Drag the point along the x-axis and watch the partial sums settle down inside the radius -- and blow up outside it.
Loading interactive…
Worked example
Radius of convergence by the ratio test
-
1
Find the radius of convergence of \displaystyle\sum_{n=0}^\infty x^n.
-
2
Apply the ratio test: \displaystyle\lim_{n\to\infty}\left|\dfrac{x^{n+1}}{x^n}\right| = |x|. The series converges when this limit is less than 1.
-
3
|x|<1 gives radius R=1. \boxed{R=1}
Loading interactive…
Worked example
A power series with infinite radius
-
1
Find the radius of convergence of \displaystyle\sum_{n=0}^\infty \dfrac{x^n}{n!} (this is e^x's own series).
-
2
Ratio test: \displaystyle\lim_{n\to\infty}\left|\dfrac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \lim_{n\to\infty}\dfrac{|x|}{n+1}=0 for every x.
-
3
The limit is 0 (less than 1) no matter how large |x| is. \boxed{R=\infty}
Loading interactive…
Where this goes next
Series solutions near the end of this course
When a differential equation's coefficients aren't constant, there often is no formula built from familiar functions — a power series, substituted directly into the equation, is sometimes the only way to construct a solution at all, and the radius found here is exactly where that constructed solution is trustworthy.
Check your understanding (counts toward your progress)
What does it mean for a power series \sum a_n(x-x_0)^n to converge at a particular value of x?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Every individual term of a convergent power series is automatically finite -- that alone isn't the definition of convergence.
Coefficients going to zero is neither necessary nor sufficient for convergence at a given x.
A power series is genuinely infinite; convergence describes what the running sum approaches, not term count.
Why
Convergence at x means the partial sums \sum_{n=0}^N a_n(x-x_0)^n approach a finite limit as N\to\infty.
Why does a series solution to a differential equation, built by substituting y=\sum a_nx^n, need its radius of convergence checked?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
A divergent series doesn't represent a real function at that x -- validity is confined to the radius.
The equation's order is fixed by its highest derivative, unrelated to any series' radius.
The same convergence requirement applies to any power series, however it was constructed.
Why
A power series only represents a genuine function within its radius of convergence -- outside it, the constructed 'solution' is meaningless.
Think about what a partial sum has to do as more terms are added.
Find the radius of convergence of \displaystyle\sum_{n=0}^\infty \dfrac{x^n}{2^n}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
The ratio test gives \lim_{n\to\infty}\left|\dfrac{x^{n+1}/2^{n+1}}{x^n/2^n}\right|=\dfrac{|x|}{2}<1, so |x|<2 and R=2.
Correct answer
Also accepted:
Why
Apply the ratio test to \left|\frac{x^{n+1}/2^{n+1}}{x^n/2^n}\right|.
Find the radius of convergence of \displaystyle\sum_{n=0}^\infty n!\,x^n.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
The ratio test gives \lim_{n\to\infty}\left|\dfrac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| = \lim_{n\to\infty}(n+1)|x| = \infty for any x\ne0, so the series converges only at x=0, giving R=0.
Correct answer
Also accepted:
Why
Apply the ratio test -- watch what happens to (n+1)|x| as n\to\infty.