Tim finds a new way to write the nth term of an infinite geometric series

Tim worked on a few infinite series like with partial sums 1/3, 4/9, 13/27, 40/81, ' which goes to ' . Don asked Tim how to get from 81 to 40 in the 4th partial sum. His first discovery was that if you add the numerator and the denominator of the previous fraction you get 40. That was fine. Was there another way to generalize the nth partial sum? Tim came up with the following, which Don had never seen before:

which he generalized to

Fine job Tim!!

In the meantime, Don did the following in Mathematica for the same series

above.

Sum[3^x/3^4,{x,0,4-1}]= 40/81

which is the 4th partial sum with n=4   that Tim did above.

Sum[3^x/3^3,{x,0,3-1}] with n=3, we get 13/27

The nth partial sum

Sum[3^x/3^n,{x,0,n-1}]

The limit of the right side above as n->∞, equals 1/2; done in Mathematica,

Limit[Sum[3^x/3^n,{x,0,n-1}],n->] = 1/2

The nth partial sum starting with 2/5 (this took Don quite a while to figure out how to write this sum and then write it as aMathematica command):

Sum[(2^x*5^(n-x))/5^n,{x,1,n}]

1/3* 2^(1 + n)* 5^(-n)*(-1 + (5/2)^n)

Don tested this answer by checking the 3rd partial sum which is 78/125.

The 3rd partial sum starting with 2/5 is

Sum[(2^x*5^(3-x))/5^3,{x,1,3}]

78/125 which is what he got above.

The limit of the series starting with 2/5 is 2/3

Limit[Sum[(2^x*5^(n-x))/5^n,{x,1,n}],n->∞] = 2/3

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