Examples for Section 8.3
Infinite Series

p. 638

Mathematica is fairly good at adding infinite series. The function to use is

Sum[a[n],  {n, 1, ∞}]

or with symbols from the Basic Input Pallette:

(∑_(n = 1)^∞) a[n]

1.

(∑_(n = 1)^k) (2/(3^(n - 1)))  // Expand

2 (((1 - ((((1/3)))^k))/(1 - (1/3))))  // Expand

(∑_(n = 1)^∞) (2/(3^(n - 1)))

2 ((1/(1 - (1/3))))

3.

(∑_(n = 1)^k)((((-1))^(n - 1))/(2^(n - 1)))  // Expand

(((1 - (((((-1)/2)))^k))/(1 - (((-1)/2)))))  // Expand

(∑_(n = 1)^∞)((((-1))^(n - 1))/(2^(n - 1)))

(1/(1 - (((-1)/2))))

11.

(∑_(n = 0)^∞)((5/(2^n)) + (1/(3^n)))

5 ((1/(1 - ((1/2))))) + (1/(1 - (1/3))) // Expand

17.

s[k_]  := 
      (∑_(n = 1)^k)
          (( 40.0  n)/((((2  n - 1))^2) (((2  n + 1))^2)))

s[10]

s[100]

Using telescoping sums:

Apart[((40.0  n)/((((2  n - 1))^2) (((2  n + 1))^2)))]

s[k_]  := 
      (∑_(n = 1)^k)((5/(((2  n - 1))^2)) - (5/(((2 n + 1))^2)))
             // N

It is clear that all but the first and last terms cancel so \(TraditionalForm
      \`s\_k = \ 5\/\((2 - 1)\)\^2 - \ 5\/\((2  k + 1)\)\^2\  -> 5\) as k -> ∞.

Limit[s[k], k -> ∞]

19.

s[k_]  := (∑_(n = 1)^k)((1/(Sqrt[n])) - (1/(Sqrt[n + 1])))

s[10]

s[100]

Using telescoping sums it is clear that (Subscript[s, k]) = 1  -  (1/(Sqrt[k + 1])) so (Subscript[s, k]) -> 1 as k -> ∞.

29.

The series (∑_(n = 0)^∞) (e^(-2 n))is a geometric series, (∑_(n = 0)^∞)((((1/(e^2))))^n), and converges to (1/(1 - (1/(e^2)))) = ((e^2)/((e^2) - 1)).

(∑_(n = 0)^∞) (E^(-2 n))

35.

The series (∑_(n = 1)^∞)(( n!)/(1000^n))diverges because the terms do not approach 0. For n  ≥ 1000 : 
\(TraditionalForm
      \`\(n!\)\/1000\^n = 
        \(n\/1000\ \((n - 1)\)\/1000 ... \)\ \(3\/1000\) \(2\/1000\) 
              1\/1000\  ≥ \ \(1000\/1000\) 999\/1000 ... \)(3/1000) (2/1000) (1/1000)

37.

The series (∑_(n = 1)^∞) ln((n/(n + 1))) can be written as a telescoping series: \(TraditionalForm
      \`∑\+\(n = 1\)\%∞\((ln(n)\  - \ ln(n + 1))\)\). The partial sums, \(TraditionalForm
      \`s\_k = \ ln(1)\  - \ ln(k + 1)\  -> \(-∞\)\)as k -> ∞ so the series diverges.

43.

The series \(TraditionalForm
      \`∑\+\(n = 0\)\%∞ 3 \((\(x - 1\)\/2)\)\^n\) is a geometric series and converges to \(TraditionalForm
      \`3 \((1\/\(1 - \((x - 1)\)\/2\))\) = 6\/\(3 - x\)\) if ((|(x - 1) | )/2)  <  1.  The last inequality is equivalent to -1 < x < 3.