How to show this sum covereges

[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.

Thanks in advance

 

[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.

 

[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).

 

[zli034]

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).

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

 

[A. Bahat]

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

 

[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.

 

[DonAntonio]

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.


*** IF...the series are positive, of course. ***


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.

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