p. 621

56.

a[n_]  :=  (((1 + (0.5/n)))^n)

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

ListPlot[z, PlotRange -> {1, 2}]

The sequence appears to converge to a number close to 1.64061. The exact number can be obtained by using L'Hopital's Rule to calculsate the limit: \(TraditionalForm
      \`lim\_\(x -> ∞\)\((1 + a\/x)\)\^x\  = \ e\^a\).

Limit[(((1 + (a/n)))^n), n -> ∞]

(E^0.5)

Limit[a[n],  n -> ∞]

Now we find an integer N such that  | (Subscript[a, n]) - (e^0.5) |   <   .01 for n  >  N.

Plot[Abs[(((1 + (0.5/x)))^x) - (E^0.5)] -  .01, {x, 1, 40}]

The graph indicates that N = 25 should be big enough:

Abs[a[25] - (E^0.5)]

Experimenting show that N = 21also works.

Abs[a[21] - (E^0.5)]

To get  | (Subscript[a, n]) - (e^0.5) |   <   .0001:

Plot[Abs[(((1 + (0.5/x)))^x) - (E^0.5)] -  .0001, {x, 1500, 3000},  
      PlotRange -> {- .0001,  .0001}]

The graph shows that N = 2100 should be good enough:

Abs[a[2100]  -  (E^0.5)]

Experimenting show that N = 2061 also works.

Abs[a[2061]  -  (E^0.5)]

60.

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

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

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

The sequence appears to approach 1. This can be verified using L'Hopital's Rule: \(TraditionalForm
      \`lim\_\(x -> ∞\)x\ \(sin(1/x)\)\  = \ 
        \(lim\_\(x -> ∞\)\(sin \((1/x)\)\)\/\(1/x\) = \ 
          \(lim\_\(x -> ∞\)\(\(cos(1/x)\) 
                  \((\(-1\)/x\^2)\)\)\/\(\(-1\)/x\^2\) = \ 
            \(cos(0)\  = \ 1\)\)\)\) Now we find an integer N such that  | (Subscript[a, n]) - 1 |  <  .01for n  >  N.

Plot[Abs[x Sin[(1/x)]  -  1]  -   .01,  {x, 1, 20}, 
      PlotRange -> {- .15,  .15}]

The graph indicates that N = 5 is good enough:

Abs[a[5]  - 1]

To get  | (Subscript[a, n]) - 1 |  <  .0001:

Plot[Abs[x Sin[(1/x)]  -  1]  -   .0001,  {x, 1, 100}, 
      PlotRange -> {- .0002,  .0002}]

The graph indicates that N = 45 is good enough:

Abs[a[45] - 1]

Experimenting show that N = 41also works.

Abs[a[41] - 1]

62.

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

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

ListPlot[z]

We know the sequence should approach 0 since by L'Hopital's Rule \(TraditionalForm
      \`lim\_\(x -> ∞\)\(ln \((x)\)\)\/x = 
        \(lim\_\(x -> ∞\)\(1/x\)\/1 = 0\)\). Now we find an integer N such that  | (Subscript[a, n]) - 0 |  <  .01for n  >  N.

Plot[Abs[Log[x]/x] -  .01, {x, 1, 100}]

Plot[Abs[Log[x]/x] -  .01, {x, 100, 1000}]

The graph indicates that N = 650 should be good enough:

Abs[a[650]]

Experimenting shows that N = 648 also works.

Abs[a[648]]

To get (Subscript[(| a), n]) |  <  .0001 we experiment with plots and find that N = 120, 000 should be good enough.

Plot[Abs[Log[x]/x] -  .0001, {x, 100000, 200000}]

Abs[a[120000]]

Further experimenting shows that N  =  116672  also works.

Abs[a[116672]]