Examples from Section 8.4
The Integral Test for Series

p. 643

3.

The series (∑_(n = 1)^∞) (n/(n + 1)) diverges since the terms do not approach 0: \(TraditionalForm
      \`lim\_\(n -> ∞\)n\/\(n + 1\) = \ 1. \) Also, using the integral test:

(∫_1^b) (x/(x + 1)) \[DifferentialD]x

The limit goes to ∞ as b -> ∞.

9.

The series \(TraditionalForm
      \`∑\+\(n = 2\)\%∞\( ln \((n)\)\)\/n\)diverges by the integral test:

(∫_2^b) Log[x]/x \[DifferentialD]x

The limit is ∞ as b -> ∞.

19.

The series \(TraditionalForm
      \`∑\_\(n = 1\)\%∞ 1\/\((ln(2))\)\^n\) diverges because it is a geometric series with r  =  (1/(ln(2))) >  1.

N[(1/Log)[2]]

Also, by the integral test

(∫_1^b) (1/Log)[2]^x \[DifferentialD]x

The term \(TraditionalForm
      \`\(ln(2)\)\^\(-b\) = \((1\/\(ln(2)\))\)\^b -> ∞\) as b -> ∞ since (1/(ln(2))) > 1.

22.

Use the integral test for the series \(TraditionalForm
      \`∑\_\(n = 1\)\%∞ 1\/\(n(1 + \(ln(n)\)\^2)\)\).

(∫_1^b) (1/(x (1 + Log[x]^2))) 
        \[DifferentialD]x

(Use the substitution u  =  1  +  ln(x).) As b -> ∞, ln(b) -> ∞ and arctan(ln(b)) -> π/2. Therefore the series converges.

23.

The series (∑_(n = 1)^∞) n sin((1/n)) diverges because the terms do not approach 0: \(TraditionalForm
      \`lim\_\(n -> ∞\ \)n\ \(sin(1\/n)\) = 
        \(lim\_\(x -> ∞\ \)\(sin(1\/x)\)\/\(1\/x\) = \ 1
          \)\) (by L'Hopital's Rule).

31.

For what values of a does the series \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\((a\/\(n + 2\) - 1\/\(n + 4\))\)\) converge?

\(TraditionalForm
      \`∑\+\(n = 1\)\%∞\((a\/\(n + 2\) - 1\/\(n + 4\))\)\  = \ 
        ∑\+\(n = 1\)\%\(∞\ 
            \)\(\((a - 1)\)\ n\  + \ 4\ a\  - \ 2\)\/\(\((n + 2)\) 
              \((n + 4)\)\)\)

If a = 1, then the series becomes  
      (∑_(n = 1)^(∞ ))(2
          /((n + 2) (n + 4))). Apply the integral test:

(∫_1^b) (2/((x + 2) (x + 4))) 
        \[DifferentialD]x

The limit ln(((b + 2)/(b + 4))) -> 1 as b -> ∞ by L'Hopital's Rule, so the series converges.

If a ≠ 0, then the integral is

(∫_1^b)
        (((a - 1) x  +  4 a  -  2)/((x + 2) 
              (x + 4))) \[DifferentialD]x

If a  >  0, then ((((b + 2))^a)/(b + 4)) -> ∞ and \(TraditionalForm
      \`ln(\((b + 2)\)\^a\/\(b + 4\)) -> ∞\) as b -> ∞, and if a  <  0, then ((((b + 2))^a)/(b + 4)) -> 0 and \(TraditionalForm
      \`ln(\((b + 2)\)\^a\/\(b + 4\)) -> \(-∞\)\) as b -> ∞. Thus the series diverges when a ≠ 0.

41.

a) From Exercise 33 we know that \(TraditionalForm
      \`\(ln(n + 1)\  ≤ 1 + 1\/2 +  ... \) + 1\/n ≤ 
        1 + ln(n)\). This implies that \(TraditionalForm
      \`\(0 < ln(n + 1) - ln(n) ≤ 1 + 1\/2 +  ... \) + 1\/n - 
          ln(n) ≤ 1\), so the sequence (Subscript[a, n]) = 1 + (1/2) +  ...  + (1/n) - ln(n)is bounded above and below.

b) The graph of y = (1/x)for n ≤ x ≤ n + 1lies above the line y = (1/(n + 1)), so \(TraditionalForm
      \`1\/\(n + 1\) < 
          ∫\_n\%\(n + 1\)\(1\/x\) dx\  = \ 
        ln(n + 1) - ln(n)\). By part a), this implies that  \(TraditionalForm
      \`a\_\(n + 1\) - a\_n = 
        \(\((\(1 + 1\/2 +  ... \) + 1\/\(n + 1\) - ln(n + 1))\) - 
            \((\(1 + 1\/2 +  ... \) + 1\/n - ln(n))\) = 
          1\/\(n + 1\) - \((ln(n + 1) - ln(n))\) < 0\)\) showing that the sequence is decreasing. A bounded decreasing sequecne must converge. The limit in this case is an important constant called Euler's constant and is usually denoted by γ.

a[n_]  :=  (∑_(k = 1)^n) (1.0/k) - Log[n]

a[100]

a[1000]

a[10000]

a[1000000]