How to show this sum covereges

1. Apr 15, 2012

zli034

$$\sum_{m=1}^N(\frac{1}{m^4}-\frac{1}{m^6})$$

My math on sum series is very rusty, can anyone show me show this sum converges?

It is not geometric series, right?

Suddenly found out it is needed to show Kolmogorov SLLN of some random varianble.

2. Apr 15, 2012

HallsofIvy

Have you done the basic algebra?
$$\frac{1}{m^4}- \frac{1}{m^6}= \frac{m^2}{m^6}- \frac{1}{s^6}= \frac{m^2- 1}{m^6}$$
Now "compare" that to $1/m^4$ which converges.

3. Apr 15, 2012

A. Bahat

Alternatively, one could note that Ʃ1/n4 and Ʃ1/n6 both converge, so Ʃ(1/n4-1/n6) converges (for good measure, it's equal to Ʃ1/n4 - Ʃ1/n6 = π4/90 - π6/945).

4. Apr 15, 2012

zli034

Why does Ʃ1/n4 not diverges? I have BS in biology, now I working with probability, only can troubleshoot math with you guys. Thanks

5. Apr 15, 2012

A. Bahat

A series of the form Ʃ1/np converges if and only if p>1. There are several ways to prove this, the most common of which is probably the integral test.

6. Apr 15, 2012

TylerH

A more general way to prove it converges is to use the theorem that if the infinite sum of f(n) converges and g(n) <= f(n) for all n, then g(n) converges.

You can use the theorem with 1/n^2, which converges. All you have to do is show that 1/n^2 > 1/n^4 - 1/n^6 for all n.

7. Apr 15, 2012

DonAntonio

Perhaps he/she doesn't know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, and if he's going to prove this he might as well prove and use

the more general answer by Bahat.

DonAntonio