Analyzing the Double Series: Finding its Sum Analytically

[Emanuel84]

Homework Statement

Determine the character of the following double series, eventually find its sum in analytical form:

$$\sum_{m,n=2}^\infty \frac{1}{m^n-1}$$2. The attempt at a solution

At first look, this series can only be either convergent or divergent (i.e. regular), since it's a positive term series.
I rewrote it in a more convenient form in order to visualize it better:

$$\sum_{m=2}^\infty \left( \sum_{n=2}^\infty \frac{1}{m^n-1} \right)$$

At this point, I thought to use the comparison test for the series within the parenthesis, being the test series the geometric series:

$$\sum_{n=2}^\infty \frac{1}{m^n}$$

which is convergent, since

$$\frac{1}{m}<1$$

Thus:

$$\sum_{n=2}^\infty \frac{1}{m^n-1}$$ converges.

Iterating this procedure, I found that the whole double series converges.

The problem, now, is to find its value analytically...any hints?

Thanks!

 

[mighty2000]

Hello there!

I would like to clarify that the character of a series refers to its properties and behavior, not its value. In this case, the series has been determined to be convergent.

To find its sum in analytical form, we can use the formula for the sum of an infinite geometric series:

S = \frac{a}{1-r}

where S is the sum, a is the first term, and r is the common ratio.

In this case, the first term is 1/m^2 and the common ratio is 1/m. Plugging these values into the formula, we get:

S = \frac{\frac{1}{m^2}}{1-\frac{1}{m}} = \frac{1}{m^2-m}

Since this is a double series, we need to sum over both the m and n terms. This can be done by using the formula for the sum of a geometric series again, but with the first term being 1/2 and the common ratio being 1/2:

\sum_{m=2}^\infty \frac{1}{m^2-m} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1

Therefore, the sum of the original double series is 1. I hope this helps! Keep up the good work in your studies.