Examples from Section 8.5
Comparison Tests for Series

p. 649

1.

The series (∑_(n = 1)^∞)(1/(2(Sqrt[n])+(n^(1/3)))) can be compared with \(TraditionalForm
      \`\(1\/3\) \(∑\+\(n = 1\)\%∞ 1\/\@n\)\) because (n^(1/3)) < (Sqrt[n]) implies (1/(3 (Sqrt[ n])))<(1/(2(Sqrt[n])+(n^(1/3)))) .  The series (∑_(n = 1)^∞) (1/(Sqrt[n])) is a p-series with p < 1, so it diverges. Hence, the original series diverges too by the Comparison Test.

3.

Compare the series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\(\( sin\^2\)(n)\)\/2\^n\) with (∑_(n = 1)^∞) (1/(2^n)) since (((sin^2)(n))/(2^n)) < (1/(2^n)). The latter series converges as a geometric series, so the former series converges too. 
The function NSum can be used to compute a numerical approximation for the series:

NSum[Sin[n]/(2^n), {n, 1, ∞}]

5.

The series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\( 2  n\)\/\(3  n - 1\)\) does not converge because the terms do not approach 0: \(TraditionalForm
      \`lim\_\(n -> ∞\)\(2  n\)\/\(3  n - 1\) = 
        2\/3 ≠ 0\).

11.

For the series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\((ln(n))\)\^2\/n\^3\), use the Limit Comparison Test with (∑_(n = 1)^∞) (1/(n^2)). The latter series converges as a p-series, and \(TraditionalForm
      \`lim\_\(n -> ∞\)a\_n\/b\_n = 
        \(lim\_\(n -> ∞\)\((
              \(\((ln(n))\)\^2\/n\^3\)/\(1\/n\^2\))\) = 
          \(lim\_\(n -> ∞\)\((ln(n))\)\^2\/n = 0\)\)\) by L'Hopital's Rule. Therefore the original series converges.

Limit[((((Log[n]))^2)/(n^3))/(1/(n^2)), n -> ∞]

NSum[((((Log[n]))^2)/(n^3)), {n, 1, ∞}]

19.

Compare the series \(TraditionalForm
      \`∑\+\(n = 2\)\%∞ 1\/\(n \@\( n\^2 - 1\)\)\) with (∑_(n = 2)^∞) (1/(n^2)) using the limit comparison test: \(TraditionalForm
      \`lim\_\(n -> ∞\)\((
            \(1\/\(n \@\( n\^2 - 1\)\)\)/\(1\/n\^2\))\) = 
        \(lim\_\(n -> ∞\)1\/\@\(1 - 1\/n\^2\) = 1\)\) . The latter series is a p-series and converges since p = 2 > 1, so the former series converges as well.

Limit[(1/(n (Sqrt[ (n^2) - 1])))/(1/(n^2)), n -> ∞]

NSum[(1/(n (Sqrt[ (n^2) - 1]))), {n, 2, ∞}]

25.

The series (∑_(n = 1)^∞) sin((1/n)) diverges because we can compare it to the harmonic series (∑_(n = 1)^∞) (1/n) which diverges: \(TraditionalForm
      \`lim\_\(n -> ∞\)\(sin(1\/n)\)/\(1\/n\) = 1\) (by L'Hopital's Rule).

Limit[Sin[(1/n)]/(1/n),  n -> ∞]

41. (Modified)

Explore the behavior of the series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞ 1\/\(100\ n\ \(\(cos\^2\)(n)\)\)\).
We first look at the partial sums:

s[k_]  :=  (∑_(n = 1)^k) (1.0/(100  n Cos[n]^2))

The decimal in the numerator is intentional so that the partial sums will be computed numerically. Mathematica cannot compute the sum exactly:

Limit[s[k], k -> ∞]

z  =  Table[s[k], {k, 1, 100}];

ListPlot[z, PlotRange -> {40, 55}]

s[100]

The partial sums appear to be approaching a limit between 49 and 50.

z  =  Table[s[k], {k, 1, 200}];

ListPlot[z, PlotRange -> {40, 55}]

s[200]

The first 200 partial sums seem to confirm that they are approaching a limit between 49 and 50. Now let us look at the first 400 terms.

z  =  Table[s[k], {k, 1, 400}];

ListPlot[z, PlotRange -> {40, 55}]

s[400]

If we look closely, we will see that noticable jumps in the partial sums appear at n = 322, n = 344, n = 366, and n = 388.

z  =  Table[s[k], {k, 300, 400}];

ListPlot[z]

The jumps occur because the corresponding values of n are close to π/2  +  k π some some integer k. Therefore, cos(n) is close to 0 and the term (1/(100 n (cos^2)(n)))is relatively large.

To find where the largest jumps will occur, we should look for numbers of the form \(TraditionalForm
      \`π/2\  + \ k\ π\  = \ \((2  k + 1)\) π/2\) that are close to integers (fractional part < .01 or > .99):

Do[If [
        FractionalPart[N[(2  k + 1) π/2 ]] <  .01 || 
          FractionalPart[N[(2  k + 1) π/2 ]] >  .99, 
        Print[ N[(2  k + 1) π/2]]], {k, 100, 300}]

A noticable jump should occur for n = 699 and n = 721:

s[697]s[698]

s[699]s[700]

s[719]s[720]

s[721]s[722]

Such jumps make it difficult to assess whether the series will ultimately converge. We do not have an easy way to estimate how small cos(n) may become (it is never zero, of course), and therefore, no easy way to estimate how large (1/(100 n (cos^2)(n))) may become. The function NSum also has trouble with this series:

NSum[(1.0/(100  n Cos[n]^2)),  {n, 1, ∞}]