Why Manipulate a Known Series?
A single familiar power series can generate many useful representations. Replacing its variable with an expression, differentiating it, or integrating it can produce series for new functions without starting from scratch. These operations are especially useful when an exact function is difficult to represent directly.
The most useful starting point is the geometric series. Its interval of convergence matters: after any manipulation, the resulting series still needs a valid interval, and endpoints must be checked rather than assumed.
Substitution: Change the Input, Then the Interval
If a series represents a function for inputs \(u\) in a particular interval, substituting an expression for \(u\) gives a representation wherever that expression lies in the original interval. For the geometric series, the condition is \(|u|<1\). After substitution, solve that inequality in the new variable. This gives the interior of the interval; boundary points still require their own convergence checks.
Worked Example: Substitute a Quadratic into the Geometric Series
Find a power series for \(f(x)=\frac{x}{1+x^2}\), and state its interval of validity.
Start with \(\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n\) for \(|u|<1\). Choose \(u=-x^2\). Then \(1-u=1+x^2\), so
The condition \(|-x^2|<1\) is \(x^2<1\), or \(-1<x<1\). Multiply by \(x\):
At \(x=1\), the series becomes \(1-1+1-1+\cdots\), whose terms do not approach zero. At \(x=-1\), it becomes \(-1+1-1+1-\cdots\), which also fails the nth-term condition. Thus the interval is \((-1,1)\).
Check: The finite geometric sum gives \(\sum_{n=0}^{N}(-1)^n x^{2n+1} =x\frac{1-(-x^2)^{N+1}}{1+x^2}\). For \(|x|<1\), the remainder term involving \((-x^2)^{N+1}\) approaches zero. The sum therefore approaches \(\frac{x}{1+x^2}\), as required.
Differentiation: A New Series Inside the Same Radius
A power series can be differentiated term by term at every point strictly inside its radius of convergence. The differentiated series has the same radius of convergence. This statement does not automatically settle convergence at either endpoint: those points must be tested in the new series.
Worked Example: Differentiate the Geometric Series
Find a power series for \(\frac{1}{(1-x)^2}\), and determine its interval of validity.
For \(|x|<1\), begin with \(\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\). Differentiate both sides within that interval:
Reindex with \(k=n-1\), so \(n=k+1\). Then
At \(x=1\), the terms are \(1,2,3,\ldots\), so they do not approach zero. At \(x=-1\), the terms are \((k+1)(-1)^k\), whose magnitudes also do not approach zero. Neither endpoint is included.
Check: The represented function is \((1-x)^{-2}\), whose derivative calculation began with \(\frac{d}{dx}(1-x)^{-1}=(1-x)^{-2}\). The first terms of the differentiated series are \(1+2x+3x^2+4x^3\), matching the derivative of \(1+x+x^2+x^3+\cdots\) term by term. The interval is \((-1,1)\), consistent with the differentiation theorem and the endpoint checks.
Integration: Another Route to New Representations
A power series can also be integrated term by term strictly inside its interval of convergence. An antiderivative of \(\sum a_nx^n\) is \(\sum \frac{a_n}{n+1}x^{n+1}+C\) there. The radius remains unchanged, but endpoint convergence may change. In particular, dividing coefficients by \(n+1\) can make an endpoint series converge even when the original series does not.
Worked Example: Integrate to Obtain the Inverse Tangent Series
Find a power series for \(\arctan x\), including its endpoint behavior.
Substitute \(u=-t^2\) into the geometric series. For \(|t|<1\),
Since \(\frac{d}{dt}\arctan t=\frac{1}{1+t^2}\) and \(\arctan 0=0\), integrate from \(0\) to \(x\), with \(|x|<1\):
Thus \(\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots\) for \(|x|<1\). At \(x=1\), the series is \(1-\frac13+\frac15-\frac17+\cdots\); it converges by the alternating-series test, since the term magnitudes decrease to zero. At \(x=-1\), it is the negative of that series, so it converges there as well.
To verify that the endpoint sums equal the function values, use the finite geometric identity \(\frac{1}{1+t^2}=\sum_{n=0}^{N}(-1)^nt^{2n}+\frac{(-1)^{N+1}t^{2N+2}}{1+t^2}\). Integrating from \(0\) to \(1\), the absolute value of the remainder integral is at most \(\int_0^1t^{2N+2}\,dt=\frac{1}{2N+3}\), which approaches zero. The partial sums therefore approach \(\int_0^1\frac{1}{1+t^2}\,dt=\arctan(1)=\frac{\pi}{4}\). Odd symmetry gives the value \(-\frac{\pi}{4}\) at \(x=-1\).
For \(|x|>1\), the terms \(\frac{(-1)^nx^{2n+1}}{2n+1}\) do not approach zero, so the series cannot converge. The interval of validity is therefore \([-1,1]\).
Check: Differentiating the series for \(|x|<1\) gives \(\sum_{n=0}^{\infty}(-1)^n x^{2n}\), the geometric representation of \(\frac{1}{1+x^2}\). The series also has value \(0\) at \(x=0\), matching \(\arctan 0\).
Why the Manipulations Work
Inside its radius of convergence, a power series converges uniformly on every closed interval strictly contained in that radius. This justifies integrating its partial sums and passing to the limit; the corresponding power-series differentiation theorem justifies differentiating term by term there. The geometric series also has a direct finite-sum explanation: for \(|u|<1\),
Substitution changes the input \(u\); differentiation and integration change the coefficients and powers. In each case, the representation follows from a known identity together with a convergence condition that permits the operation.
Common Pitfall: Endpoints Do Not Come for Free
A common error is to solve the strict convergence condition and then include equality cases automatically. For example, substituting \(u=x^2\) into the geometric series gives
The condition gives \(-1<x<1\), not \(-1\le x\le1\). A mistaken argument might say that \(x^2\le1\) at the endpoints and therefore include them. But at either endpoint the series becomes \(1+1+1+\cdots\), which diverges. The function is also undefined at \(x=\pm1\). The correct interval is \((-1,1)\).
For any transformed series, first find the range where the original power-series operation is justified. Then test each boundary point in the resulting series. A familiar function may be defined at a boundary even though its derived series diverges there, or the series may converge at an endpoint that was not covered by the interior argument.
Check Your Understanding
For each question, state the relevant interval and explain any endpoint decision.
- Use substitution in the geometric series to obtain a power series for \(\frac{1}{1+x}\). What is its interval of convergence?
- Differentiate \(\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) to write a series for \(\frac{1}{(1-x)^2}\). Why must the endpoints be checked separately?
- Integrate the series for \(\frac{1}{1-x}\) from \(0\) to \(x\) to obtain a series for \(-\ln(1-x)\), valid initially for \(|x|<1\).
- For the inverse tangent series, what does the term test show when \(|x|>1\)?
- Why does a convergence condition such as \(|u|<1\) not by itself determine whether a transformed series converges at its boundary points?
