Reciprocal Series of Positive Integer Factors: Convergence & Sum

[Kenshin]

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.


this is my first time writing here. i hope someone can help me with this question.

 

[robert Ihnot]

Try working with (1+1/2+1/4+1/8...)(1+1/3+1/9+1/27+++++++) See if that covers the sum above...

 

[M98Ranger]

Hello! Thank you for sharing this question. This is a very interesting series and it's great that you are looking into its convergence and sum.

To prove that this series converges, we can use the comparison test. This test states that if we have two series, one of which is always greater than the other, and the smaller series converges, then the larger series also converges.

In this case, we can compare our given series to the series 1+1/2+1/4+1/8+... which is a geometric series with a common ratio of 1/2. This series converges to 2.

Now, let's look at the series given to us. Every term in this series is either equal to or less than the corresponding term in the geometric series. This is because every term in our series has either 2 or 3 as its only prime factors, while the terms in the geometric series have all powers of 2 as their prime factors.

Therefore, our series is always less than or equal to the geometric series and thus, it also converges to 2.

To find the sum of this series, we can use the formula for the sum of a geometric series.

S = a/(1-r), where S is the sum, a is the first term and r is the common ratio.

In our series, a = 1 and r = 1/2.

Therefore, S = 1/(1-1/2) = 1/(1/2) = 2.

Hence, the sum of this series is 2.

I hope this helps! Let me know if you have any further questions. Good luck with your studies!