Examples for Section 8.2
Calculating Limits of Sequences

p. 628

1.

a[n_]  :=  2  +  (0.1^n);

z  =  Table[a[n],  {n, 1, 10}]

Limit[a[n], n -> ∞]

5.

a[n_]  :=  N[((1 - 5  (n^4))/((n^4) + 8  (n^3)))];

z  =  Table[a[n],  {n, 1, 50}]

Limit[a[n], n -> ∞]

Apart[((1 - 5  (n^4))/((n^4) + 8  (n^3)))]

All terms except -5 go to 0 as n -> ∞.

11.

a[n_]  :=  N[(((n + 1)/(2  n))) (1  -  (1/n))];

z  =  Table[a[n],  {n, 1, 20}]

Limit[a[n], n -> ∞]

The first factor is ((n + 1)/(2  n))  =  (1/2) + (1/(2  n))which goes to (1/2)and the second factor 1 - (1/n) goes to 1, so the product goes to (1/2) 1  =  (1/2).

37.

a[n_]  :=  N[((n!)/(n^n))];

z  =  Table[a[n], {n, 1, 10}]

ListPlot[z,  PlotRange -> {{0, 10}, {0, 1}}]

It looks like (Subscript[a, n]) -> 0 as n -> ∞. To see this, we exapnd out n! and (n^n)as products and use the fact that (k/n) ≤ 1 for 1 ≤ k ≤ n to get: \(TraditionalForm
      \`0 ≤ \(n!\)\/n\^n = \ 
        \(n\/n\ \(\(n - 1\)\/n\) \(n - 2\)\/n ... \) \(3\/n\) \(2\/n\) 1\/n\  
          ≤ 1\/n\). Since (1/n) -> 0, the Sandwich Theorem implies that ((n!)/(n^n)) -> 0.

77.

Solve the equation cos(x)  +  x  =  0 using Picard's Method.

Plot[Cos[x] + x, {x, -2 π, 2  π}]

To use Picard's method correctly in this example, we should choose f(x)  =  -cos(x)  -  x (the roots of f(x) are the same as the solutions of cos(x) + x  =  0), and then \(TraditionalForm
      \`g(x)\  = \ 
        \(f(x) + x = \(\(-\(cos(x)\)\) - x + x = \(-\(cos(x)\)\)\)\)\). We create a sequence (Subscript[x, n]) starting with a first approximation to the solution (which we get from the above graph): (Subscript[x, 0]) =  -1 . The next terms in the sequence are obtained by \(TraditionalForm
      \`x\_n = \ \(g(x\_\(n - 1\)) = \(-\(cos(x\_\(n - 1\))\)\)\)\).

g[x_]  :=  N[-Cos[x]];

\(x[0]\  = \ \(-1\); 
x[n_]\  := \ g[x[n - 1]]\)

z  =  Table[x[n],  {n, 1, 20}]

ListPlot[z,  PlotRange -> {-1, 0}]

The sequence converges to a number close to -0.739184 which gives a good approximation to the solution of cos(x) + x  =  0:

Cos[-0.739184]+(-0.739184)