Examples for Section 8.8
Power Series

p. 671

3.

Find the radius and interval of convergence of the series \(TraditionalForm
      \`∑\+\(n = 0\)\%∞\(\((\(-1\))\)\^n\) 
          \((4  x + 1)\)\^n\).

Since this is a geometric series, we know it will converge absolutely if  | 4  x + 1 |  < 1 (to the sum (1/(1 + (4  x + 1)))) and diverge if  | 4  x + 1 |  ≥ 1. (The Root Test is based on the geometetric series and gives the same conclusion.) The inequality  | 4  x + 1 |  < 1 is equivalent to  | x - (-(1/4)) |  < (1/4) which shows that the center is -(1/4)and the radius of convergence is (1/4). The corresponding interval is -(1/2) < x  < 0.

p10  =  
      (∑_(n = 0)^10)(((-1))^n) (((4  x + 1))^n)

Plot[{(1/(2 + 4  x)), p10}, {x, -1/2, 0}, 
      PlotRange -> {0, 10}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[1, 0, 0]}]

9.

Find the radius and interval of convergence of the series (∑_(n = 1)^∞) ((x^n)/(n (Sqrt[ n]))).

Apply the Root Test: \(TraditionalForm
      \`\( | x\^n\/\(n \@ n\)\( | \^\(1/n\)\)\) = 
        1\/n\^\(3/\((2  n)\)\) | x |  -> \( | x | \)\) since \(TraditionalForm
      \`lim\_\(n -> ∞\)n\^\(3/\((2  n)\)\) = 
        \(lim\_\(n -> ∞\)e\^\(3 \( ln(n)\)/\((2  n)\)\) = 
          \(e\^0 = 1\)\)\). Therefore the series converges absolutely for  | x |  < 1and diverges for  | x |  > 1. The center is 0 and the radius of convergence is 1. At either endpoint x = ±1 the series converges absolutely since \(TraditionalForm
      \`∑\+\(n = 1\)\%∞ 1\/\(n \@ n\) = 
        ∑\+\(n = 1\)\%∞ 1\/n\^\(3/2\)\) is a p-series with p > 1.

p10  =  (∑_(n = 1)^10) ((x^n)/(n (Sqrt[ n])))

Plot[p10, {x, -1/2, 0}, PlotStyle -> RGBColor[1, 0, 0]]

13.

Find the radius and interval of convergence of the series \(TraditionalForm
      \`∑\+\(n = 0\)\%∞ x\^\(2  n + 1\)\/\(n!\)\).

Apply the Ratio Test: \(TraditionalForm
      \`\( | a\_\(n + 1\)/a\_n | \) = 
        \(\( | \(x\^\(2  n + 3\)\/\(\((n + 1)\)!\)\)/
              \(x\^\(2  n + 1\)\/\(n!\)\) | \) = 
          \( | x\( | \^2\)\)\/\((n + 1)\) -> 0\)\) for any x. Therefore the series converges absolutely for all x. The center is 0 and the radius of convergence is ∞. We shall soon see that \(TraditionalForm
      \`e\^u = \ ∑\+\(n = 0\)\%∞ u\^n\/\(n!\)\)for all u (see also Problem 42) and the given series can be written in terms of this: \(TraditionalForm
      \`∑\+\(n = 0\)\%∞ x\^\(2  n + 1\)\/\(n!\) = 
        \(x \(∑\+\(n = 0\)\%∞\((x\^2)\)\^n\/\(n!\)\) = 
          x\ e\^\(x\^2\)\)\).

p10  =  (∑_(n = 1)^10) ((x^(2  n + 1))/(n!))

Plot[{x (E^((x^2))), p10},  {x, -2, 2}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[1, 0, 0]}]

33.

Find the radius and interval of convergence of the series \(TraditionalForm
      \`∑\+\(n = 0\)\%∞\((x - 1)\)\^\(2  n\)\/4\^n\) and find the sum of the series in this interval.

The series is geometric: \(TraditionalForm
      \`∑\+\(n = 0\)\%∞\((x - 1)\)\^\(2  n\)\/4\^n\)= (∑_(n = 0)^∞) (r^n) where r = ((((x - 1))^2)/4). Therefore it converges absolutely for  | ((((x - 1))^2)/4) |  < 1 and diverges for \(TraditionalForm
      \`\( | \((x - 1)\)\^2\/4 | \( ≥ 1\)\)\). The first inequality is equivalent to \(TraditionalForm\`\( | x - 1\( | \^2\)\( < 4\)\)\) or  | x - 1 |  < 2 which shows that the center is 1 and the radius is 2. The interval of convergence is thus -1 < x < 3. Usingthe fromula for a geometric series we find that \(TraditionalForm
      \`∑\+\(n = 0\)\%∞\((x - 1)\)\^\(2  n\)\/4\^n = 
        \(1\/\(1 - \((x - 1)\)\^2\/4\) = 
          \(4\/\(4 - \((x - 1)\)\^2\) = 4\/\(3 + 2  x - x\^2\)\)\)\)

p10  =  (∑_(n = 0)^10)((((x - 1))^(2  n))/(4^n))

Plot[{(4/(3 + 2  x - (x^2))), p10},  {x, -1, 3}, 
      PlotRange -> {0, 10}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[1, 0, 0]}]

41.

The series
	\(TraditionalForm
      \`sin(x)\  = \ 
        x\  - \ x\^3\/\(3!\) + \ x\^5\/\(5!\) - x\^7\/\(7!\) + x\^9\/\(9!\) - 
            x\^11\/\(11!\) + \  ... \)
converges for all x. Integrating term by term gives:
	 \(TraditionalForm
      \`\(-\(cos(x)\)\) = 
        \(∫\(sin(x)\) dx\  = \ 
          C\  + \(1\/2\) x\^2 - \ x\^4\/\(4!\) + x\^6\/\(6!\) - 
              x\^8\/\(8!\) + x\^10\/\(10!\) + \  ... \)\)
Since cos (0) = 1, we see that C = -1and we obtain:
	\(TraditionalForm
      \`cos(x)\  = \ 
        1\  - \ x\^2\/\(2!\) + \ x\^4\/\(4!\) - x\^6\/\(6!\) + x\^8\/\(8!\) - 
            x\^10\/\(10!\) + \  ... \)
which also converges for all x. We could also take the derivative of sin(x)

 If we plug in 2  x for x in the above epxression for sin(x) we get:
	\(TraditionalForm
      \`sin(2  x)\  = \ 
        2  x\  - \ \((2  x)\)\^3\/\(3!\) + \ \((2  x)\)\^5\/\(5!\) - 
            \((2  x)\)\^7\/\(7!\) + \((2  x)\)\^9\/\(9!\) - 
            \((2  x)\)\^11\/\(11!\) + \  ... \)
We know that sin(2  x)  =  2  sin(x) cos(x) and if we multiply the above series for sin(x) and cos(x) we get:
2  sin(x) cos(x) = 2(x  -  ((x^3)/(3!)) +  ((x^5)/(5!)) - 
                    ((x^7)/(7!)) + ((x^9)/(9!)) - ((x^11)/(11!)) +   ... )(1  -  ((x^2)/(2!)) +  ((x^4)/(4!)) - ((x^6)/(6!)) + 
                      ((x^8)/(8!)) - ((x^10)/(10!)) +   .. .)

Expand[
      2 (x - ((x^3)/(3!)) + ((x^5)/(5!)) - ((x^7)/(7!)) + ((x^9)/(9!)) - 
            ((x^11)/(11!))) 
        (1 - ((x^2)/(2!)) + ((x^4)/(4!)) - ((x^6)/(6!)) + ((x^8)/(8!)) - 
            ((x^10)/(10!)))]

Expand[ 
      2  x  -  ((((2  x))^3)/(3!)) +  ((((2  x))^5)/(5!)) - 
        ((((2  x))^7)/(7!)) + ((((2  x))^9)/(9!)) - 
        ((((2  x))^11)/(11!))]

Thus, multiplying out like polynomials the first six terms of the series for 2  sin(x) cos(x)  yields the first six terms series for sin(2  x).

46. (Modified)

Find the sum of the series (∑_(n = 1)^∞) ((n^3)/(3^n)).

Start with \(TraditionalForm
      \`1\/\(1 - x\) = ∑\+\(n = 0\)\%∞ x\^n\), take its derivative, \(TraditionalForm
      \`1\/\((1 - x)\)\^2 = \ 
        ∑\+\(n = 1\)\%∞ n\ x\^\(n - 1\)\), and multiply by x to get(x/(((1 - x))^2)) =  (∑_(n = 1)^∞) n (x^n). Now take the derivative and multiply by x again: \(TraditionalForm
      \`x\ \(d\/\(d\ x\)\) \((x\/\((1 - x)\)\^2)\) = 
        x \( d\/\(d\ x\)\) \(∑\+\(n = 1\)\%∞ n\ x\^n\)\). This simplifies to \(TraditionalForm
      \`\(x(x + 1)\)\/\((1 - x)\)\^3 = \ 
        ∑\+\(n = 1\)\%∞ n\^2\ x\^n\).

x D[(x/(((1 - x))^2)), x] // Simplify

Repeat these steps one more time: \(TraditionalForm
      \`x\ \(d\/\(d\ x\)\) \((\(x(x + 1)\)\/\((1 - x)\)\^3)\) = 
        x \( d\/\(d\ x\)\) \(∑\+\(n = 1\)\%∞ n\^2\ x\^n\)\) which simplifies to \(TraditionalForm
      \`\(x(x\^2 + 4  x + 1)\)\/\((1 - x)\)\^4 = \ 
        ∑\+\(n = 1\)\%∞ n\^3\ x\^n\).

x D[((x(x + 1))/(((1 - x))^3)), x] // Simplify

This converges absolutely for  | x |  < 1 since the orginal geometric series converges there. Plugging in x = (1/3)gives
	\(TraditionalForm
      \`∑\+\(n = 1\)\%∞ n\^3\/3\^n = 
        \(\(\(1\/3\) \((\((1\/3)\)\^2 + 4  1\/3 + 1)\)\)\/\((1\/3 - 1)\)\^4 = 
          33\/8\)\).
Mathematica also knows how to sum this series:

(∑_(n = 1)^∞) ((n^3)/(3^n))