Examples for Section 8.1
Limits of Sequences

p. 619

3.

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

Table[a[n],  {n, 1, 4}]

9.

Clear[a]

\(a[1]\  := \ 2; 

    a[n_]\  := \ \(\(\((\(-1\))\)\^\(n + 1\)\) a[n - 1]\)\/2\)

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

55.

a)

Clear[a];

a[n_]  :=  N[(n^(1/n))]

z  =  Table[a[n],  {n, 1, 25}]

ListPlot[z]

The sequence appears to be bounded below by 1, and in fact (Subscript[a, n]) -> 1 as n -> ∞.
b) Find N such that  | (Subscript[a, n]) - 1 |   <   .01 for n  >  N. Repeat for .0001.
Experimenting:

a[200]  -  1

a[400]  -  1

a[600]  -  1

a[700] - 1

So, N  =  700 will work.
A plot can also help us find where (n^(1/n)) - 1  <   .01 iff (n^(1/n)) - 1.01 < 0:

Plot[(n^(1/n)) - 1.01, {n, 100, 800}]

To get  (n^(1/n)) - 1  <   .0001 we can look at plots again:

Plot[(n^(1/n)) - 1.0001, {n, 100000, 200000}]

So N  =  130, 000 will work.

a[130000] - 1

57

a)

Clear[a]

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

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

ListPlot[z]

The sequence appears to be increasing rapidly towards 1.25.
b) Find N such that  | (Subscript[a, n]) - 1.25 |   <   .01 for n  >  N.  Repeat for .0001.

z = Table[a[n] - 1.25,  {n, 1, 25}]

The convergence is so fast, N = 4 will work for .01 and N  =  6 will work for .0001

59.

a)

Clear[a]

a[n_]  :=  N[Sin[n]];

z  =  Table[a[n],  {n, 1, 25}]

l = ListPlot[z]

The points appear random, but they are all onthe graph of y  =  sin(x):

p = Plot[Sin[x], {x, 1, 25}, PlotStyle -> Hue[0.3]];

Show[p, l]

No matter how many terms we consider, the sequence does not appear to approach any number.

z  =  Table[a[n],  {n, 1, 100}];

l = ListPlot[z]

p = 
      Plot[Sin[x], {x, 1, 100}, PlotPoints -> 50, 
        PlotStyle -> Hue[0.3]];

Show[p, l]

61.

a)

Clear[a];

a[n_]  :=  N[Sin[n]/n]

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

l = ListPlot[z,  PlotRange -> {- .5,  .5}]

p = Plot[Sin[x]/x, {x, 1, 50}, PlotStyle -> Hue[0.3]];

Show[p, l,  PlotRange -> {- .5,  .5}]

The limit appears to be 0. This is clear since \(TraditionalForm
      \`\( | \(sin(n)\)\/n | \( ≤ 1\/n\)\) -> 0\) as n -> ∞.
b) Find N such that (Subscript[(| a), n]) |   <   .01 for n  >  N.  Repeat for .0001.
Again, since  | ((sin(n))/n) |   ≤ (1/n), we need only make (1/n) <   .01 or .0001. Therefore, \(TraditionalForm\`N = \(1\/ .01 = 100\)\) works for the first, and \(TraditionalForm\`N\  = \ \(1\/ .0001 = \ 10\), 000\) works for the second.

a[100]

a[10000]