COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
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This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 4  SEQUENCES, JULIA and MANDELBROT SETS, and POWER SERIES

c = q = S = t = x = y = z = 
                    0; 
Clear[a, absZn, c, cx, cy, dot, i, juliaC, k, lim, limX, limY, limZ, 
        list, mandelbrotC, n, newton, Pn, q, Qn, R, s, S, Sn, set, sum, sumN, 
        sumX, sumY, sumZ, t, table, x, Xn, y, Yn, z, Z, 
        Zn]; 
Needs["Graphics`ComplexMap`"]; 
Needs["Graphics`Colors`"];

Section 4.1	Sequences and Series

Section 4.2	Julia and Mandelbrot Sets

Section 4.3  Geometric Series and Convergence Theorems

We begin this section by studying series of the form (∑_(n = 0)^∞) (z^n), which is called a geometric series, one of the most important series in mathematics.

Theorem 4.11,  Page 144.  (Geometric Series)  If  |z|<1,  the series  (∑_(n = 0)^∞) (z^n)  converges to  f (z) = (1/(1  -  z)).  That is, if |z|<1 then  

			(∑_(n = 0)^∞) (z^n)= 1 + z + (z^2) +  ...  + (z^k) +  ...  =(1/(1  -  z)).  

If  |z|≥1,  the series diverges.

Proof of Theorem 4.11,  see text Page 144.

Corollary 4.2,  Page 145.   If  |z|>1,  the series  (∑_(n = 1)^∞) (z^(-n))  converges to  f (z) = (1/(z -  1)).  That is, if |z|>1 then  

			(∑_(n = 1)^∞) (z^(-n))=(z^(-1)) + (z^(-2)) +  ...  + (z^(-4)) +  ...  =(1/(z -  1)),  
or equivalently,
			-(∑_(n = 1)^∞) (z^(-n))=(z^(-1)) - (z^(-2)) -  ...  - (z^(-4)) -  ...  =(1/(1 - z)).  

If  |z|≤1,  the series diverges.

Proof of Corollary 4.2,  see text Page 145.

Corollary 4.3,  Page 146.   If  z ≠ 1, then for all n,  

			(1/(1  -  z)) = 1 + z + (z^2) +  ...  + (z^(n - 1)) + 
          ((z^n)/(1  -  z)).

Proof of Corollary 4.3,  see text Page 146.

Example 4.13,  Page 146.  Show that  (∑_(n = 0)^∞)((((1  -  i))^n)/(2^n)) = 
        1 - i.

Solution 4.13.

We can use the definition of convergence of a series and find the limit of the partial sums.

Remove[S, z]; 
Clear[k, lim, n, sum, Z]; 
(Subscript[z, k])_  =  ((((1  -  I))^k)/(2^k)); 
(Subscript[S, n])_  = (∑_(k = 0)^n)((((1 - I))^k)/(2^k)); 
sum  = (∑_(k = 0)^∞)((((1 - I))^k)/(2^k)); 
lim = Limit[(Subscript[S, n]),n->∞]; 
\(Print[\*""\!\(z\_k\) = "", z\_k];\) 
Print[""]; 
\(Print[\*""(Subscript[S, n]) = (∑_(k = 0)^n)((((1 - I))^k)/(2^k)) = "", S\_n];\) 
Print[""]; 
\(Print[\*""(lim_(n -> ∞)) "", \*""(Subscript[S, n]) = "", lim];\) 
\(Print[\*""(∑_(k = 0)^∞)((((1 - I))^k)/(2^k)) = "", sum];\) 
Null

Or we can see that this is an infinite geometric series with ratio  Z = (1/2) - (i/2).

Z  =  
        ((Subscript[z, k + 1])/(Subscript[z, k])); 
\(\(Print[\*"" \!\(z\_k\) = "", z\_k];\)\ \)
\(\(Print[\*""\!\(z\_\(k + 1\)\) = "", z\_\(k + 1\)];\)\ \)
Print[""]; 
\(\(Print[\*""  Z = \!\(z\_\(k + 1\)\/z\_k\) = "", 
        Z];\)\ \)

The sum of the infinite geometric series is now found.

Print["          Z = ", Z]; 
Print["        |Z| = ", Abs[Z]]; 
Print["     Is |Z| < 1 ?  ", Abs[Z] <  1]; 
Print[" "]; 
\(\(Print[\*""      (1/(1 - Z)) = (1/(1 - ((1/2) - (I/2))))""];\)\ \)
\(\(Print[\*""      \!\(1\/\(1 - Z\)\) = "", \ 
        1\/\(1 - \((1\/2 - \(\(\ \)\(I\)\)\/2)\)\)];\)\ \)
\(\(Print[\*""(∑_(k = 0)^∞)((((1 - I))^k)/(2^k)) = "", ∑\+\(k = 0\)\%∞\((1 - I)\)\^k\/2\^k];\)\ \)

The series of absolute values converges, therefore the series converges.

Use Mathematica to construct some of the partial sums of the infinite series.

(Subscript[S, n])_  =  (∑_(k = 0)^n)((((1 - I))^k)/(2^k)); 
table = 
        Table[FullSimplify[(Subscript[S, n])], {n, 0, 10}]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(k = 0)^∞)((((1 - I))^k)/(2^k)) = "", ∑\+\(k = 0\)\%∞\((1 - I)\)\^k\/2\^k];\)\ \)

Use Mathematica to plot some of the partial sums of the infinite series.

set  =   Table[{Re[(Subscript[S, n])], Im[(Subscript[S, n])]}, {n, 0, 
            15}]; 
ListPlot[set, 
        Prolog -> {PointSize[0.021], 
            Blue}, 
PlotRange -> {{0, 
              1.55}, {-1.3, 
              0.0}}, 
AxesLabel -> {"x", "y"}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[0, 1.6, 0.2], 
            Range[-1.4, 0, 0.2]}]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(k = 0)^∞)((((1 - I))^k)/(2^k)) = "", sum];\)\ \)

We see that the sum of the infinite geometric series  \(∑\+\(n = 0\)\%∞\((1\  - \ i)\)\^n\/2\^n\ \ \ is\ \ \ 1 - i\).

Example 4.14,  Page 146.  Evaluate  \(∑\+\(n = 3\)\%∞\(\(\ \)\(i\^n\)\)\/2\^n\).

Solution 4.14.

We can use the definition of convergence of a series and find the limit of the partial sums.

Remove[S, z]; 
Clear[k, lim, n, sum, Z]; 
(Subscript[z, k])_  = (( (I^k))/(2^k)); 
(Subscript[S, n])_ = (∑_(k = 3)^∞) ((I^k)/(2^k)); 
sum = (∑_(k = 3)^∞)(( (I^k))/(2^k)); 
lim = Limit[(Subscript[S, n]),n->∞]; 
\(Print[\*""\!\(z\_k\) = "", z\_k];\) 
Print[""]; 
\(Print[\*""(Subscript[S, n]) = (∑_(k = 3)^n)((I^k)/(2^k)) = "", S\_n];\) 
\(Print[\*""      (lim_(n -> ∞)) "", \*""(Subscript[S, n]) = "", lim];\) 
\(Print[\*""     (∑_(k = 3)^∞)((I^k)/(2^k)) = "", sum];\)

Or we can see that this is an infinite geometric series with ratio  Z = (i/2).

Z  =  
        ((Subscript[z, k + 1])/(Subscript[z, k])); 
\(\(Print[\*"" \!\(z\_k\) = "", z\_k];\)\ \)
\(\(Print[\*""\!\(z\_\(k + 1\)\) = "", z\_\(k + 1\)];\)\ \)
Print[""]; 
\(\(Print[\*""  Z = \!\(z\_\(k + 1\)\/z\_k\) = "", 
        Z];\)\ \)

The sum of the infinite geometric series is now found.

a  =  (Subscript[z, 3]); 
\(Print[\*""        a = ((I/2))^3 = "", a];\) 
Print[""]; 
Print["        Z = ", Z]; 
Print["      |Z| = ", Abs[Z]]; 
Print["   Is |Z| < 1 ?  ", Abs[Z] <  1]; 
Print[" "]; 
\(Print[\*""     (a/(1 - Z)) = (((((I/2)))^3)/(1 - (I/2)))""];\) 
\(Print[\*""     \!\(a\/\(1 - Z\)\) = "", a\/\(1 - Z\)];\) 
Print[""]; 
Print["    (∑_(k = 3)^∞)((I^k)/(2^k)) = ",(∑_(k = 3)^∞) ((I^k)/(2^k))];

Use Mathematica to construct some of the partial sums of the infinite series.

(Subscript[S, n])_ = (∑_(k = 3)^n) ((I^k)/(2^k)); 
table = Table[FullSimplify[(Subscript[S, n])], {n, 0, 10}]; 
table = Append[table, "..."]; 
\(Print[\*""{\!\(S\_n\)} = "", table];\) 
Print[""]; 
\(Print[\*""{\!\(S\_n\)} = "", N[table]];\) 
Print[""]; 
Print["(∑_(k = 3)^∞)((I^k)/(2^k)) = ",sum];

Use Mathematica to plot some of the partial sums of the infinite series.

set  =   Table[{Re[(Subscript[S, n])], Im[(Subscript[S, n])]}, {n, 0, 
              10}]; 
\(ListPlot[set, 
          Prolog -> {PointSize[0.02], 
              Magenta}, 
PlotRange -> {{\(-0.05\), 
                0.55}, {\(-0.55\), 
                0.05}}, 
AxesLabel -> {"x", "y"}, 
          AspectRatio -> 
            1, 
Ticks -> {Range[0, 0.5, 0.1], 
              Range[\(-0.5\), 0, 0.1]}];\) 
\(Print[\*""{\!\(S\_n\)} = "", N[table]];\) 
Print[""]; 
Print["(∑_(k = 3)^∞)((I^k)/(2^k)) = ",sum];

We see that the sum of the infinite geometric series  (∑_(n = 3)^∞)(( (i^n))/(2^n))  is  (1/20)-(i/10).

Remark 4.3. The equality given by this example illustrates an important point when evaluating a geometric series whose beginning index is other than zero. The value of (∑_(n = r)^∞) (z^n) equals((z^r)/(1  -  z)). If we think of z as the "ratio'' by which a given term of the series is multiplied to generate successive terms, we see that the sum of a geometric series equals ((first term)/(1  -  ratio)), provided  |ratio|<1.

The geometric series is used in the proof of theorem 4.12 and is 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,  Page 147.  (d'Alembert's Ratio Test)  If  (∑_(n = 1)^∞) (Subscript[ζ, n])  is a complex series with the property that  

			\(lim\+\(n\  -> \ ∞\)\ \(\(|\)\(ζ\_\(n + 1\)\)\(|\)\)\/\(\(|\)\(ζ\_n\)\(|\)\) = L\),  

then the series converges absolutely if  L < 1  and diverges if  L > 1.

Proof of Theorem 4.12,  see text Page 147.

Example 4.15,  Page 147.  Show that  \(\(\(\ \)\(∑\+\(n = 0\)\%∞\((1\  - \ i)\)\^n\/\(n!\)\)\)\)  converges.

Solution 4.15.

Enter the formula for the terms in the series.

Remove[z]; 
Clear[L, n, sum, r]; 
(Subscript[z, n])_  =  (((((1  -  I))^n) )/(n!)); 
\(\(Print[\*""\!\(z\_n\) = "", 
        z\_n];\)\ \)

Use d'Alembert's ratio test. First, find the ratio of consecutive terms.

\(
\(Print[\*""         \!\(z\_\(n + 1\)\) = "", 
        z\_\(n + 1\)];\)\ \)
\(\(Print[\*""          \!\(z\_n\) = "", z\_n];\)\ \)
r  =  ((Subscript[z, n + 1])/(Subscript[z, n])); 
Print[""]; 
\(\(Print[\*""        \!\(z\_\(n + 1\)\/z\_n\) = "", 
        r];\)\ \)
r  =  FullSimplify[r]; 
\(\(Print[\*""        \!\(z\_\(n + 1\)\/z\_n\) = "", 
        r];\)\ \)
Print[""]; 
\(\(Print[\*""       |\!\(z\_\(n + 1\)\/z\_n\)| = "", 
        Abs[Numerator[r]]\/Denominator[r]];\)\ \)
Print[""]; 
L  =  Limit[ r ,  n -> ∞]; 
\(\(Print[\*""L = (lim_(n -> ∞)) |((Subscript[z, n + 1])/(Subscript[z, n]))| = "", L];\)\ \)

Since L < 1, the series converges.   

We can use Mathematica to find the sum of series.

sum   =   (∑_(n = 0)^∞)((((1 - I))^n)/(n!)); 
\(\(Print[\*""(∑_(n = 0)^∞)((((1 - I))^n)/(n!)) = "", sum];\)\ \)

Use Mathematica to construct some of the partial sums of the infinite series.

(Subscript[S, n])_  =  (∑_(k = 0)^n)((((1 - I))^k)/(k!)); 
table = 
        Table[FullSimplify[(Subscript[S, n])], {n, 0, 10}]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""{\!\(S\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(n = 0)^∞)((((1 - I))^n)/(n!)) = "", sum];\)\ \)

Use Mathematica to plot some of the partial sums of the infinite series.

set  =   Table[{Re[(Subscript[S, n])], Im[(Subscript[S, n])]}, {n, 0, 
            10}]; 
\(\(ListPlot[set, 
        Prolog -> {PointSize[0.022], 
            Orange}, 
PlotRange -> {{\(-0.05\), 
              2.35}, {\(-2.35\), 
              0.05}}, 
AxesLabel -> {"x", "y"}, 
        AspectRatio -> 
          1, 
Ticks -> {Range[0, 2.5, 0.5], 
            Range[\(-2.5\), 0, 0.5]}];\)\ \)
\(\(Print[\*""{\!\(S\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""(∑_(n = 0)^∞)((((1 - I))^n)/(n!)) = "", sum];\)\ \)

We see that the infinite series  \(\(\(\ \)\(∑\+\(n = 0\)\%∞\((1\  - \ i)\)\^n\/\(n!\)\)\)\)  converges and that its sum is  (e^(1 - i)).

Example 4.16,  Page 147.  Show that  \(∑\+\(n = 0\)\%∞\((z\  - \ i)\)\^n\/2\^n\) converges for all  z  in the disk  |z - i|< 2.

Solution 4.16.

Enter the formula for the terms in the series.

Remove[z]; 
Clear[L, n, sum, r, Z]; 
(Subscript[z, n])_  =  ((((z  -  I))^n)/(2^n)); 
\(\(Print[\*""\!\(z\_n\) = "", 
        z\_n];\)\ \)

Use d'Alembert's ratio test. First, find the ratio of consecutive terms.

r  =  
        ((Subscript[z, n + 1])/(Subscript[z, n])); 
L  =  
        Limit[ r ,  n -> ∞]; 
\(\(Print[\*""         \!\(z\_\(n + 1\)\) = "", 
        z\_\(n + 1\)];\)\ \)
\(\(Print[\*""          \!\(z\_n\) = "", 
        z\_n];\)\ \)
Print[""]; 
\(\(Print[\*""        \!\(z\_\(n + 1\)\/z\_n\) = "", 
        r];\)\ \)
Print[""]; 
\(\(Print[\*""L = (lim_(n -> ∞)) |((Subscript[z, n + 1])/(Subscript[z, n]))| = "", Abs[L]];\)\ \)

When  L < 1, the series will converge.  Solve |((z  -  i)/2)|<1  and obtain the disk  |z - i|< 2.  

Use Mathematica to find the sum of the infinite series.

f[z_] = (∑_(n = 0)^∞)((((z - I))^n)/(2^n)); 
\(Print[\*""f[z] = (∑_(n = 0)^∞)((((z - I))^n)/(2^n)) = "", f[z]];\)

We can investigate the convergence by plotting several partial sums of this series.

Since convergence will be more rapid in a smaller disk  |z - i|< r,  the following plot will be a smaller disk with  r = 1.

Clear[F, G, m]; 
F[z_]  =  f[z + I]; 
S[z_, n_]=(∑_(n = 0)^n)((((z - I))^n)/(2^n)); 
\(For\ [m = 3, 
      m ≤ 5, \(m++\), 
G[z_]\  = \ S[z + I, m]; \ 

      PolarMap[G\ , {0, 1.0, 0.2}, {0, 
          2  π, π\/12}, 
PlotRange -> {{0, 
              2}, {\(-0.7\), 0.7}}, 
        AspectRatio -> 
          7\/10, 
Ticks -> {Range[0, 2, 1], 
            Range[\(-1\), 1, 1\/2]}, AxesLabel -> {"u", "u"}, 
        PlotStyle -> {Green, 
            DarkGreen}]; \ 
Print["S[z,", 
        m, "] = ", S[z, m]];\ \ ]\)
PolarMap[
          F , {0, 1.0, 0.2}, {0, 
            2  π, (π/12)}, 
PlotRange -> {{0, 
                2}, {-0.7, 0.7}}, 
          AspectRatio -> 
            (7/10), 
Ticks -> {Range[0, 2, 1], 
              Range[-1, 1, (1/2)]}, AxesLabel -> {"u", "v"}, 
          PlotStyle -> {Red, Magenta}]; 
Print["f[z] = ", f[z]];

Convergence will be slower in a larger disk  |z - i|< r,  the following plot will be from a larger disk with  r = 1.6.

Clear[F, G, j, m]; 
F[z_]  =  f[z + I]; 
S[z_, n_]=(∑_(n = 0)^n)((((z - I))^n)/(2^n)); 
For [j = 3, 
      j ≤ 5, j++, 
m = 
        5*j;  
G[z_]  =  
        S[z + I, m];  
PolarMap[
        G , {0, 1.6, 0.2}, {0, 
          2  π, (π/12)}, 
PlotRange -> {{0, 
              2}, {-0.7, 0.7}}, 
        AspectRatio -> 
          (7/10), 
Ticks -> {Range[0, 2, 1], 
            Range[-1, 1, (1/2)]}, AxesLabel -> {"u", "u"}, 
        PlotStyle -> {Cyan, 
            Blue}];  
Print["S[z,", m, "] = ", 
        S[z, m]];  ]
PolarMap[
          F , {0, 1.6, 0.2}, {0, 
            2  π, (π/12)}, 
PlotRange -> {{0, 
                2}, {-0.7, 0.7}}, 
          AspectRatio -> 
            (7/10), 
Ticks -> {Range[0, 2, 1], 
              Range[-1, 1, (1/2)]}, AxesLabel -> {"u", "v"}, 
          PlotStyle -> {Red, Magenta}]; 
Print["f[z] = ", f[z]];

We see that the sequence of functions {S[x,n]} is converging to f[z].

funs = 
        Table[{(Subscript[s, m])[z], "=", S[z, m]}, {m, 1, 
            5}]; 
funs = 
        Append[funs, {"...", "", "..."}]; 
funs = 
        Append[funs, {"f[z]", "=", 
            f[z]}]; 
TableForm[funs]

Our next result, known as the root test, is slightly more powerful than the ratio test. Before we state this test, we need to discuss a rather sophisticated idea that it uses---the limit supremum.

Definition 4.9 (Limit Supremum), Page 148.  Let  {(Subscript[t, n])}  be a sequence of positive real numbers.  The limit supremum of the sequence  (denoted by (Underscript[(lim sup), n  ->  ∞]) (Subscript[t, n]))  is the smallest real number  L  with the property that for any  ϵ > 0  there are at most finitely many terms in the sequence that are larger than  L + ϵ.  If there is no such number  L,  then we set  (Underscript[(lim sup), n  ->  ∞]) (Subscript[t, n]) = ∞.

Example 4.17,  Page 148.  The limit supremum of the sequence {(Subscript[t, n])} = {4.1, 5.1, 4.01, 5.01, 4.001, 5.001,  ... }  is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 5.

Solution 4.17.

In this case the even terms  (Subscript[t, 2 k]) = 4 + (10^(-1 - k))  tend to the limit 4 and the odd terms  (Subscript[t, 2 k + 1]) = 5 + (10^(-1 - k))  tend to the limit 5.

Remove[t]; 
Clear[limsup, k, n, table]; 
(Subscript[t, even, k_]) = 
        4 + (1/(10^(k + 1))); 
(Subscript[t, odd, k_]) = 5 + (1/(10^(k + 1))); 
lim1 = 
        Limit[(Subscript[t, even, k]), 
          k -> ∞]; 
lim2 = Limit[(Subscript[t, odd, k]), k -> ∞]; 
limsup = Max[lim1, lim2]; 
table = 
        Flatten[Table[{(Subscript[t, even, k]), (Subscript[t, odd, k])}, {k, 0, 
              4}]]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""  \!\(t\_\(2  k\)\) = ("", 
        t\_\(even, k\), ")"];\)\ \)
\(\(Print[\*"" \!\(t\_\(2  k + 1\)\) = ("", 
        t\_\(odd, k\)\ , ")"];\)\ \)
Print[""]; 
\(\(Print[\*"" (lim_(k -> ∞)) (Subscript[t, 2  k]) = (lim_(k -> ∞)) ("", t\_\(even, k\)\ , ") = ", 
        lim1];\)\ \)
\(\(Print[\*""(lim_(k -> ∞)) (Subscript[t, 2  k + 1]) = (lim_(k -> ∞)) ("", 
        t\_\(odd, k\)\ , ") = ", lim2];\)\ \)
Print[""]; 
\(\(Print[\*""(Underscript[limsup, n -> ∞]) (Subscript[t, n]) = Max["", lim1, ",", lim2, "]"];\)\ \)
\(\(Print[\*""(Underscript[limsup, n -> ∞]) (Subscript[t, n]) = "", limsup];\)\ \)

The limit superior is the largest limit point of a subsequence of  {(Subscript[t, n])}.  

We can use Mathematica to graph the sequence and look for the largest limit point of a subsequence.

pts0 = 
        Table[{Green, Point[N[{2  k, (Subscript[t, even, 2  k])}]]}, {k, 0, 
            10}]; 
pts1 = 
        Table[{Magenta, Point[N[{2  k + 1, (Subscript[t, odd, 2  k + 1])}]]}, {k, 0, 
            10}]; 
pts = Union[pts0, pts1]; 
\(\(Show[Graphics[pts], Axes -> Automatic, 
        Prolog -> PointSize[0.02], PlotRange -> {{0, 20}, {0, 6}}, 
        AxesLabel -> {"k", \*""(Subscript[t, k])""}, 
Ticks -> {Range[0, 20, 2], 
            Range[0, 6, 1]}];\)\ \)
\(\(Print[\*""{\!\(t\_n\)} = "", 
        N[table]];\)\ \)

We see that the limit supremum of the sequence {(Subscript[t, n])} = {4.1, 5.1, 4.01, 5.01, 4.001, 5.001,  ... }  is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 5.

Example 4.18,  Page 148.  The limit supremum of the sequence {(Subscript[t, n])} = {1, 2, 3, 1, 2, 3, 1, 2, 3,  ... }  is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 3.

Solution 4.18.

In this case there are only three different values for the terms in the sequence.

Clear[k]; 
table = 
        Table[1 + Mod[k, 3], {k, 0, 15}]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""(Underscript[limsup, n -> ∞]) (Subscript[t, n]) = Max[1,2,3]""];\)\ \)
\(\(Print[\*""(Underscript[limsup, n -> ∞]) (Subscript[t, n]) = "", Max[1, 2, 3]];\)\ \)

The limit superior is the largest limit point of a subsequence of  {(Subscript[t, n])}.  

We can use Mathematica to graph the sequence and look for the largest limit point of a subsequence.

pts0 = 
        Table[{Green, Point[N[{3  k, 1}]]}, {k, 0, 
            10}]; 
pts1 = 
        Table[{Blue, Point[N[{3  k + 1, 2}]]}, {k, 0, 
            10}]; 
pts2 = 
        Table[{Magenta, Point[N[{3  k + 2, 3}]]}, {k, 0, 
            10}]; 
pts = Union[pts0, pts1, pts2]; 
\(\(Show[Graphics[pts], Axes -> Automatic, 
        Prolog -> PointSize[0.02], 
        PlotRange -> {{0, 30}, {0, 3.5}}, 
        AxesLabel -> {"k", \*""(Subscript[t, k])""}, 
Ticks -> {Range[0, 30, 3], 
            Range[0, 4, 1]}];\)\ \)
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)

We see that the limit supremum of the sequence {(Subscript[t, n])} = {1, 2, 3, 1, 2, 3, 1, 2, 3,  ... }  is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 3.

Example 4.19,  Page 148.  The limit supremum of the Fibonacci sequence {(Subscript[t, n])}= {1, 1, 2, 3, 5, 8, 13, 21, 34,  ... } is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = ∞.

Solution 4.19.

In this case the sequence has  ∞  as its limit, and hence the limit supremum  is also  ∞.

Remove[t]; 
Clear[n]; 
(Subscript[t, n])_ = ((-(Sqrt[5]))/5) (2^(-1 - n)) (((1 - (Sqrt[5])))^(n + 1))  + (((((2/(1 + (Sqrt[5])))))^(-1 - n))/(Sqrt[5])); 
table = 
        Table[Simplify[(Subscript[t, n])], {n, 0, 12}]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)
\(\(Print[\*""\!\(t\_n\) = "", 
        t\_n];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) \!\(t\_n\) = "", 
        Limit[t\_n, n -> ∞]];\)\ \)
Print[""]; 
\(\(Print[\*""(Underscript[limsup, n -> ∞]) (Subscript[t, n]) = "", Limit[t\_n, n -> ∞]];\)\ \)

Use Mathematica to plot some of the terms in the sequence.

set  =   Table[{n, Simplify[(Subscript[t, n])]}, {n, 1, 
            12}]; 
\(\(ListPlot[set, Prolog -> {PointSize[0.02], Blue}, 
        PlotRange -> {{0, 12}, {0, 250}}, 
        AxesLabel -> {"n", \*""(Subscript[t, n])""}, 
Ticks -> {Range[0, 12, 1], 
            Range[0, 250, 50]}];\)\ \)
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)

We see that the limit supremum of the Fibonacci sequence {(Subscript[t, n])} = {1, 1, 2, 3, 5, 8, 13, 21, 34,  ... } is  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = ∞.

The limit supremum is a powerful idea because the limit supremum of a sequence always exists, which is not true for the ordinary limit. However, our next example illustrates the fact that if the limit of a sequence does exist, it will be the same as the limit supremum.

Example 4.29,  Page 149.  The sequence  {(Subscript[t, n])} = {1 + (1/n)}= \({2, 1.5, 1.3 \( \(3\)\(\ \)\)\&_, 1.25, 1.2,  ... }\)  has  (Underscript[(lim sup), n -> ∞]) (Subscript[t, n]) = 1.

Solution 4.20.

In this case the sequence has  1  as its limit, and hence the limit supremum  is also  1.

Remove[t]; 
Clear[n]; 
(Subscript[t, n])_ = 1 + (1/n); 
table = Table[(Subscript[t, n]), {n, 1, 20}]; 
table = Append[table, "..."]; 
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)
Print[""]; 
\(\(Print[\*""{\!\(t\_n\)} = "", 
        N[table]];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(t\_n\) = "", 
        t\_n];\)\ \)
Print[""]; 
\(\(Print[\*""\!\(lim\+\(n -> ∞\)\) \!\(t\_n\) = "", 
        Limit[t\_n, n -> ∞]];\)\ \)
Print[""]; 
\(\(Print[\*""(Underscript[limsup, n -> ∞]) (Subscript[t, n]) = "", Limit[t\_n, n -> ∞]];\)\ \)

Use Mathematica to plot some of the terms in the sequence.

set  =   Table[{n, Simplify[(Subscript[t, n])]}, {n, 1, 
            20}]; 
\(\(ListPlot[set, Prolog -> {PointSize[0.02], Green}, 
        PlotRange -> {{0, 20}, {0, 2}}, 
        AxesLabel -> {"n", \*""(Subscript[t, n])""}, 
Ticks -> {Range[0, 20, 2], 
            Range[0, 2, 1]}];\)\ \)
\(\(Print[\*""{\!\(t\_n\)} = "", 
        table];\)\ \)

In this case the sequence has  1  as its limit, and hence the limit supremum  is also  1.

Theorem 4.13,  Page 149.  (Root Test)  Suppose that the series  (∑_(n = 1)^∞) (Subscript[ζ, n]),  has  (Underscript[(lim sup), n  ->  ∞])  ((\(\(|\)\(\(ζ\_n\)\(|\)\)\))^(1/n))=L.  
Then the series is absolutely convergent if  L < 1  and divergent if  L > 1.

Proof of Theorem 4.10,  Page 149. We leave verification of this as an exercise.

Note that in applying either Theorems 4.12 or 4.13, if L=1 the convergence or divergence of the series is unknown, and further analysis is required to determine the true state of affairs.

Section 4.1  Exercises for Geometric Series and Convergence Theorems
See textbook page 150.

Section 4.4	Power Series Functions

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