The sum of function where n starts at 1 and goes to infinity?

Question

[4^n] + [8^n]
----------------
13^n

I thought it would be simply 12/13 but no.... apparently it is not

I am aware aqs how it is possible to get any of the terms I need whether it be 10 or 19876, I need to know how to find the overall sum this function expresses until infinity.

Answer 1

If you call your formula a(n), then you can break it up into two parts:
a(n) = 4^n/13^n + 8^n/13^n = (4/13)^n + (8/13)^n
You can sum up the two parts separately.

The sum of a geometric series (going from 1 to infinity)
sum[x^n] = x + x^2 + x^3 + ...
is equal to x/(1-x) if x is between -1 and +1.

Therefore,
sum [(4/13)^n]
= (4/13) + (4/13)^2 + (4/13)^3 + ...
=(4/13)/(1 - 4/13) = (4/13)/(9/13)
= 4/9

Also,
sum[(8/13)^n]
= (8/13)/(1 - 8/13) = (8/13)/(5/13)
= 8/5

So the summation of your original formula is
4/9 + 8/5 = 20/45 + 72/45 = 92/45
= 2.04444444444....

Some previous answers got the wrong answer because of using 1/(1-x) for the sum of a geometric series. This would be the correct formula if the sum went from 0 to infinity, but when you go from 1 to infinity, the sum is x/(1-x).

Answer 2

Put all the terms as the same number raised to a power:

4^n = 2^2n, 8^n = 2^3n

For 13^n, find x such that 2^x = 13^n

xlog2 = nlog13

x = n*log13/log2 x = 3.7004397n

The expression becomes [2^2n + 2^3n]/2^3.7004397n

2^n(2-3.7004397) + 2^n(3-3.7004397)

2^(-1.7004397)n + 2^(-0.7004397)n

The sum is then ∑0.30769231^n + ∑0.6153846^n

This is the sum of two geometric series, ∑r^n, and the value is 1/(1-r). For the first term r = 0.3077 and the sum is 1.44444; for the second term, r = 0.6155 and the sum is 2.600000.

The full sum value is then 1.4444444 + 2.60000 = 4.04444

Answer 3

for n = 0, 1st term = 2
for n = 1, 2nd term = 12/13
for n = 2, 3rd term = 80/169
for n = 3, 4th term = 253/165
for n = 4, 4th term = 16/105
for n = 5, 5th term = 81/890
and so on...
if you continue the process, the answer will gradually approach to zero, the sum is equal to 4 and 2/45 or 4.044444444

http://www.mathalino.com