COMPLEX ANALYSIS: Maple Worksheets, 2009
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
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CHAPTER 4 SEQUENCES, JULIA and MANDELBROT SETS, and Power Series

Section 4.3 Geometric Series and Convergence Theorems

We begin this section by studying series of the form , which is called a geometric series , one of the most important series in mathematics.

Theorem 4.11, (Geometric series) If , the series converges to .

That is, if , then = = .

If , then the series diverges.

Corollary 4.2 If , the series converges to .

That is, if , then = = .

If , then the series diverges.

Corollary 4.3 If , then for all

.

Example 4.13, Page 146. Show that ,

> n:='n': Zn:='Zn': z:='z':
Z := n -> (1-I)^n/2^n:
z[n] = Z(n);

This is a geometric series with ratio:

> R := Z(n+1)/Z(n):
`R = `, z[n+1]/z[n] = R;
R := simplify(R):
`R = `, z[n+1]/z[n] = R;
`|R| = `, abs(z[n+1]/z[n]) = abs(R);
`|R| = `, abs(z[n+1]/z[n]) = evalf(abs(R));

Since , the sum is given by the calculation:

> S:='S':
S := 1/(1-R):
`R = `, z[n+1]/z[n] = R;
`S = 1/(1 - R) ` = S;

Or we can use Maple to sum the series directly.

> n:='n': S:='S': Zn:='Zn': z:='z':
Zn := (1 - I)^n/2^n:
z[n] = Zn;
S := sum(Zn,n=0..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;

Error, (in convert/hypergeom) cannot evaluate boolean: 1/2*2^(1/2) < 1

Example 4.14, Page 146. Evaluate .

> n:='n': Z:='Z': Zn:='Zn': z:='z':
Z := n -> I^n/2^n:
z[n] = Z(n);

This is a geometric series with ratio:

> R := Z(n+1)/Z(n):
`R = `, z[n+1]/z[n] = R;
R := simplify(R):
`R = `, z[n+1]/z[n] = R;
`|R| = `, abs(z[n+1]/z[n]) = abs(R);
`|R| = `, abs(z[n+1]/z[n]) = evalf(abs(R));

Since |R| < 1, the sum is given by the calculation:

> S:='S':
S := 1/(1-R) - 1 - R - R^2:
`R = `, z[n+1]/z[n] = R;
`S = 1/(1 - R) - 1 - R - R^2 ` = S;

Or we can use Maple to sum the series directly.

> n:='n': S:='S': Zn:='Zn': z:='z':
Zn := I^n/2^n:
z[n] = Zn;
S := sum(Zn, n=3..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;

The geometric series is used in the proof of the following theorem, known as the ratio test . It is one of the most commonly used tests for determining the convergence or divergence of series. The proof is similar to the one used for real series, and is left for the reader to establish.

Theorem 4.12 (d'Alembert's ratio test)

If is a complex series with the property that ,

then the series is absolutely convergent if , and divergent if .

Example 4.15, Page 147. Show that converges.

> L:='L': n:='n': R:='R': Z:='Z': Zn:='Zn': z:='z':
Z := n -> (1 - I)^n / n!:
z[n] = Z(n);
R := Z(n+1)/Z(n):
z[n+1]/z[n] = R;
R := simplify(R):
z[n+1]/z[n] = R;
L := limit(Z(n+1)/Z(n) , n=infinity):
limit(z[n+1]/z[n], n=infinity) = L;

Since , the series converges. Let us use Maple to find the sum.

> n:='n': S:='S': Zn:='Zn': z:='z':
Zn := (1-I)^n/n!:
z[n] = Zn;
S := sum(Zn, n=0..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;
Sum(z[n], n=1..infinity) = evalc(S);

Example 4.16, Page 147. Show that converges for all in the disk .

> L:='L': n:='n': R:='R': Z:='Z': Zn:='Zn': z:='z':
Z := n -> (z-I)^n/2^n:


z[n] = Z(n);
R := Z(n+1)/Z(n):
z[n+1]/z[n] = R;
R := simplify(R):
z[n+1]/z[n] = R;
L := limit(Z(n+1)/Z(n) , n=infinity):
`L = `, limit(z[n+1]/z[n], n=infinity) = L;
`|L| ` = abs(L);

Hence , now solve and obtain .
Therefore, the series converges the disk .

Let us use Maple to find the sum.

> n:='n': S:='S': Zn:='Zn': z:='z':
Zn := (z-I)^n/2^n:
z[n] = Zn;
S := sum(Zn, n=0..infinity):
`Find the sum of the series:`;
Sum(z[n], n=1..infinity) = Sum(Zn, n=1..infinity);
Sum(z[n], n=1..infinity) = S;

End of Section 4.3.