Physics 374
Fall 2005
Prof. E. F. Redish

MP 44 Solution

     First, we want to investigate how quickly (∑_(n = 0)^∞)((((1/3)))^n)  converges.  Using the Table function to construct the relevant table (where the j-th entry is the sum of the first j terms of the series) is made very easy by the sum construct in the pallette.

ListPlot[Table[(∑_(n = 0)^j)((((1/3)))^n), {j, 1, 30, 1}], 
      PlotRange -> {0, 2}]

We see that the partial sum has already reached about 95% of its limiting value after including only about the first 5 terms.  Compare this with (∑_(n = 0)^∞)((((2/3)))^n)  :

ListPlot[Table[(∑_(n = 0)^j)((((2/3)))^n), {j, 1, 30, 1}], 
      PlotRange -> {0, 4}]

Which requires about the first 10 terms to reach 95% of its limiting value.  For  (∑_(n = 0)^∞)((((9/10)))^n)  :

ListPlot[Table[(∑_(n = 0)^j)((((9/10)))^n), {j, 1, 50, 1}], 
      PlotRange -> {0, 12}]

...and this last one apparently requires about the first 30 terms to reach 95% of its limiting value.  
      The observation that (∑_(n = 0)^∞)(((x))^n) converges faster for smaller x values makes sense if we take a look at some of the individual terms in the sum.  In order for the sum to be approaching its limiting value, each individual term must start heading towards zero as n gets bigger and bigger.  (Though that's not a sufficient condition for the convergence of any series.  (∑_(n = 1)^∞) (1/n) diverges even though (1/n) -> 0 as n->∞).  Anyways, let's take a look at the n = 5 term for our three example series:  ((((1/3)))^5) = .0041, ((((2/3)))^5) = .13, and ((((9/10)))^5)= .59.  We see that for the smallest "x" (i.e. one-third), the fifth term of the sum is already quite small compared to the sum's limiting value of 1.5 (.0041 is only 0.3% of 1.5).  Adding on further, higher-n terms won't change the partial sum value too much since the n = 6 term will be smaller still.  For the middle x value, we see that the fifth term is a somewhat larger contributer to the limiting value of that sum (.13 is 4.3% of 3).  For the large x, we see that the fifth term of the sum accounts for 5.9% of the limiting value.  Certainly you'd need to raise nine-tenths to a relatively large value of n before the individual terms become a fraction of a percent of the sum's limiting value.

(Solution by T .  Bing)