Convergence of a series based on reciprocals of prime factors of 2 & 3

[Kenshin]

i don't even know where to start and i hate series. if someone could get me stared that would be great help. thanks

The terms of this series are reciprocals of positive integers whose only prime factors are 2s and 3s:

1+1/2+1/3+1/4+1/6+1/8+1/9+1/12+...

Show that this series converges and find its sum.

 

[arildno]

Think of the infinite series as a sum of the folllowing subseries:
1. The subseries lacking "2" as a factor
2. The subseries with "2" as a single factor.
3. The subseries with 2 as a double factor (i.e, 2^2)
And so on..
We have the following sums:
1: 3/2
2: 1/2*3/2
3:1/4*3/2
and so on.
Hence, the total sum is 3.

Note that your infinite series is simply the Cauchy-product:
$$\sum_{n=0}^{\infty}\frac{1}{2^{n}}\sum_{m=0}^{\infty}\frac{1}{3^{m}}$$

End note:
Don't double post.

 

[wais]

To start, we can rewrite the series as:

1 + 1/2 + 1/3 + (1/2)(1/2) + (1/2)(1/3) + (1/3)(1/2) + (1/2)(1/2)(1/2) + (1/2)(1/2)(1/3) + (1/3)(1/2)(1/2) + (1/3)(1/3) + ...

We can see that the terms can be grouped into subsets based on the number of 2s and 3s in the denominator. For example, the first subset contains terms with no 2s or 3s in the denominator (1), the second subset contains terms with one 2 in the denominator (1/2, 1/3), the third subset contains terms with two 2s in the denominator (1/4, 1/6, 1/8), and so on.

Now, we can rewrite each subset as a geometric series and find its sum. For the first subset, the sum is simply 1. For the second subset, the sum is 1/2 + 1/3 = 1/2 + 1/2 = 1. For the third subset, the sum is 1/4 + 1/6 + 1/8 = 1/4 + 1/4 = 1/2. And so on.

Therefore, the sum of the entire series is the sum of all the subsets, which is 1 + 1 + 1/2 + 1/2 + 1/2 + ... = 2 + 1/2 + 1/4 + 1/8 + ...

This is a geometric series with a common ratio of 1/2 and the sum can be found using the formula S = a/(1-r), where a is the first term and r is the common ratio. So, the sum is 2/(1-1/2) = 4.

Thus, the series converges to a sum of 4.