CM 4.4 Infinite Series MrG 2011.1114

3675 days ago by calcpage123

[(1/2^n).n() for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.500000000000000, 0.250000000000000, 0.125000000000000, 0.0625000000000000, 0.0312500000000000, 0.0156250000000000, 0.00781250000000000, 0.00390625000000000, 0.00195312500000000\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.500000000000000, 0.250000000000000, 0.125000000000000, 0.0625000000000000, 0.0312500000000000, 0.0156250000000000, 0.00781250000000000, 0.00390625000000000, 0.00195312500000000\right]
sum([(1/2^n).n() for n in range(1,100)]) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}1.00000000000000
\newcommand{\Bold}[1]{\mathbf{#1}}1.00000000000000
[(1/n).n() for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.500000000000000, 0.333333333333333, 0.250000000000000, 0.200000000000000, 0.166666666666667, 0.142857142857143, 0.125000000000000, 0.111111111111111\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.500000000000000, 0.333333333333333, 0.250000000000000, 0.200000000000000, 0.166666666666667, 0.142857142857143, 0.125000000000000, 0.111111111111111\right]
sum([(1/n).n() for n in range(1,1000)]) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}7.48447086055034
\newcommand{\Bold}[1]{\mathbf{#1}}7.48447086055034
[(1/n^2).n() for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.250000000000000, 0.111111111111111, 0.0625000000000000, 0.0400000000000000, 0.0277777777777778, 0.0204081632653061, 0.0156250000000000, 0.0123456790123457\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.250000000000000, 0.111111111111111, 0.0625000000000000, 0.0400000000000000, 0.0277777777777778, 0.0204081632653061, 0.0156250000000000, 0.0123456790123457\right]
for i in [10,100,1000,10000,100000,1000000]: print sum([(1/n^2).n() for n in range(1,i)]) 
        
1.53976773116654
1.63488390018489
1.64393356668156
1.64483406184807
1.53976773116654
1.63488390018489
1.64393356668156
1.64483406184807
(pi^2/6).n() 
        
1.64493406684823
1.64493406684823
for i in range(1000,10000,1000): print sum([(1/n^2).n() for n in range(1,i)]) 
        
1.64393356668156
1.64443394182740
1.64460067795317
1.64468403559563
1.64473404684690
1.64476738629190
1.64479119950080
1.64480905903540
1.64482294956405
1.64393356668156
1.64443394182740
1.64460067795317
1.64468403559563
1.64473404684690
1.64476738629190
1.64479119950080
1.64480905903540
1.64482294956405
#1) [1-1/(n+1) for n in range(10,100,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[\frac{10}{11}, \frac{20}{21}, \frac{30}{31}, \frac{40}{41}, \frac{50}{51}, \frac{60}{61}, \frac{70}{71}, \frac{80}{81}, \frac{90}{91}\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[\frac{10}{11}, \frac{20}{21}, \frac{30}{31}, \frac{40}{41}, \frac{50}{51}, \frac{60}{61}, \frac{70}{71}, \frac{80}{81}, \frac{90}{91}\right]
[1-1/(n+1).n() for n in range(1000,10000,1000)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.999000999000999, 0.999500249875062, 0.999666777740753, 0.999750062484379, 0.999800039992002, 0.999833361106482, 0.999857163262391, 0.999875015623047, 0.999888901233196\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.999000999000999, 0.999500249875062, 0.999666777740753, 0.999750062484379, 0.999800039992002, 0.999833361106482, 0.999857163262391, 0.999875015623047, 0.999888901233196\right]
var('n') limit(1-1/(n+1),n=infinity) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}1
\newcommand{\Bold}[1]{\mathbf{#1}}1
#2) no limit((1+n)/(100*n),n=infinity) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{100}
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{100}
[(1+n)/(100*n).n() for n in range(10000,100000,10000)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.0100010000000000, 0.0100005000000000, 0.0100003333333333, 0.0100002500000000, 0.0100002000000000, 0.0100001666666667, 0.0100001428571429, 0.0100001250000000, 0.0100001111111111\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[0.0100010000000000, 0.0100005000000000, 0.0100003333333333, 0.0100002500000000, 0.0100002000000000, 0.0100001666666667, 0.0100001428571429, 0.0100001250000000, 0.0100001111111111\right]
#3) the sum is pi! 
       
#4) 1+r+r^2+r^3+...=1/(1-r) only if abs(r)<1 [3/2^n for n in range(10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[3, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \frac{3}{32}, \frac{3}{64}, \frac{3}{128}, \frac{3}{256}, \frac{3}{512}\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[3, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \frac{3}{32}, \frac{3}{64}, \frac{3}{128}, \frac{3}{256}, \frac{3}{512}\right]
[(3/2^n).n() for n in range(10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[3.00000000000000, 1.50000000000000, 0.750000000000000, 0.375000000000000, 0.187500000000000, 0.0937500000000000, 0.0468750000000000, 0.0234375000000000, 0.0117187500000000, 0.00585937500000000\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[3.00000000000000, 1.50000000000000, 0.750000000000000, 0.375000000000000, 0.187500000000000, 0.0937500000000000, 0.0468750000000000, 0.0234375000000000, 0.0117187500000000, 0.00585937500000000\right]
for i in range(2,60,2): print sum([(3/2^n).n() for n in range(i)]) 
        
4.50000000000000
5.62500000000000
5.90625000000000
5.97656250000000
5.99414062500000
5.99853515625000
5.99963378906250
5.99990844726562
5.99997711181641
5.99999427795410
5.99999856948853
5.99999964237213
5.99999991059303
5.99999997764826
5.99999999441206
5.99999999860302
5.99999999965075
5.99999999991269
5.99999999997817
5.99999999999454
5.99999999999864
5.99999999999966
5.99999999999991
5.99999999999998
5.99999999999999
6.00000000000000
6.00000000000000
6.00000000000000
6.00000000000000
4.50000000000000
5.62500000000000
5.90625000000000
5.97656250000000
5.99414062500000
5.99853515625000
5.99963378906250
5.99990844726562
5.99997711181641
5.99999427795410
5.99999856948853
5.99999964237213
5.99999991059303
5.99999997764826
5.99999999441206
5.99999999860302
5.99999999965075
5.99999999991269
5.99999999997817
5.99999999999454
5.99999999999864
5.99999999999966
5.99999999999991
5.99999999999998
5.99999999999999
6.00000000000000
6.00000000000000
6.00000000000000
6.00000000000000
limit(sum([(3/2^n).n() for n in range(i)]),i=infinity) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}6.0
\newcommand{\Bold}[1]{\mathbf{#1}}6.0
#5) series diverges [1/(2*n-1) for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \frac{1}{17}\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \frac{1}{9}, \frac{1}{11}, \frac{1}{13}, \frac{1}{15}, \frac{1}{17}\right]
[(1/(2*n-1)).n() for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.333333333333333, 0.200000000000000, 0.142857142857143, 0.111111111111111, 0.0909090909090909, 0.0769230769230769, 0.0666666666666667, 0.0588235294117647\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.333333333333333, 0.200000000000000, 0.142857142857143, 0.111111111111111, 0.0909090909090909, 0.0769230769230769, 0.0666666666666667, 0.0588235294117647\right]
for i in range(10000,100000,10000): sum([(1/(2*n-1)).n() for n in range(1,i)]) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}5.58687519670698
\newcommand{\Bold}[1]{\mathbf{#1}}5.93347378870581
\newcommand{\Bold}[1]{\mathbf{#1}}6.13621467641153
\newcommand{\Bold}[1]{\mathbf{#1}}6.28005987941549
\newcommand{\Bold}[1]{\mathbf{#1}}6.39163415512419
\newcommand{\Bold}[1]{\mathbf{#1}}6.48279660021582
\newcommand{\Bold}[1]{\mathbf{#1}}6.55987313062252
\newcommand{\Bold}[1]{\mathbf{#1}}6.62663971980284
\newcommand{\Bold}[1]{\mathbf{#1}}6.68553193208300
\newcommand{\Bold}[1]{\mathbf{#1}}5.58687519670698
\newcommand{\Bold}[1]{\mathbf{#1}}5.93347378870581
\newcommand{\Bold}[1]{\mathbf{#1}}6.13621467641153
\newcommand{\Bold}[1]{\mathbf{#1}}6.28005987941549
\newcommand{\Bold}[1]{\mathbf{#1}}6.39163415512419
\newcommand{\Bold}[1]{\mathbf{#1}}6.48279660021582
\newcommand{\Bold}[1]{\mathbf{#1}}6.55987313062252
\newcommand{\Bold}[1]{\mathbf{#1}}6.62663971980284
\newcommand{\Bold}[1]{\mathbf{#1}}6.68553193208300
#limit(sum([(1/(2*n-1)).n() for n in range(1,i)]),i=infinity) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}5.53418938278
\newcommand{\Bold}[1]{\mathbf{#1}}5.53418938278
#6) [1/n^3 for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \frac{1}{216}, \frac{1}{343}, \frac{1}{512}, \frac{1}{729}\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1, \frac{1}{8}, \frac{1}{27}, \frac{1}{64}, \frac{1}{125}, \frac{1}{216}, \frac{1}{343}, \frac{1}{512}, \frac{1}{729}\right]
[(1/n^3).n() for n in range(1,10)] 
        
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.125000000000000, 0.0370370370370370, 0.0156250000000000, 0.00800000000000000, 0.00462962962962963, 0.00291545189504373, 0.00195312500000000, 0.00137174211248285\right]
\newcommand{\Bold}[1]{\mathbf{#1}}\left[1.00000000000000, 0.125000000000000, 0.0370370370370370, 0.0156250000000000, 0.00800000000000000, 0.00462962962962963, 0.00291545189504373, 0.00195312500000000, 0.00137174211248285\right]
for i in range(100000,1000000,100000): sum([(1/n^3).n() for n in range(1,i)]) 
        
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690310973
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690314853
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690310973
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690314853
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032
\newcommand{\Bold}[1]{\mathbf{#1}}1.20205690315032