# Determine if the functions converge or diverge?I dont know how to do these! Please help by explaining so I can understand! Thanks! Example: (5n^4)+1/((150,348n^3)+999) and how you could do...

Determine if the functions converge or diverge?

I dont know how to do these! Please help by explaining so I can understand! Thanks!

Example:

(5n^4)+1/((150,348n^3)+999)

and how you could do something like this:

2,1, 2/3, 1/2, 2/5

Thanks!

### 3 Answers | Add Yours

In math we describe a function Converge or Diverge based on its ending behavior as x approches positive or negative infinity.

If a fuctions approches inf. as x approches positive or negative infinity, then the function is diverge.

On the other hand, If a function approaches a certain value as x apporaches positive or negative inf, then the function is Converge.

In your example :

(5n^4)+1/((150,348n^3)+999)

as n approaches inf., the function approches 0 , so the function is converge to 0.

and the series:

2,1, 2/3, 1/2, 2/5,....

the series is clearly approching 0 as n values approaches inf. then the series is Converge to 0.

wouldnt the first one be diverge while the second would be converge?

n^4/n^3 = n which would go to infinity therefore it is diverge, just as neela said

the second would be converge as hala178 sid because it keeps going smaller and smaller in decimals finally approaching 0

A sries like u1+u2+u3+.....un+..... is said to be convergent if its partial sum Sn = u1+u2+u3+u4+....un has a finite limit as n approaches infinity.

A necessary condtion for this is that the nth term should have limit 0.

In this case the nth term un = (5n^4)+1/((150,348n^3)+999) is

Lt (5n^4) +Lt (1/((150,348n^3)+999) as n--> inf.

= inf+0 = inf. So the series diverges;

Even if you write the nth term as (5n^4+1)/((150,348n^3)+999). The limit of this behave like 5n^4/150348n^3 as n-->infinity. Or like

(5/150348) n which approaches infinity as n--> infinity.

So the series diverge.

b)

To detrmine the nature of (convergent or otherwise) of the series 2, 1, 2/3, 2/5.... The series is rewritten as:

2/1, 2/2, 2/3, 2/4, 2/5, 2/6,........2/n.......... Or

So the nth term of this series is 2/n. Which could be compared with popular series 1, 1/2, 1/3,1/4,1/5, etc. which is divergent in the sense Su(1/n) for n=1 to inf approach infinite.

Since each term of the given series is 2 times the latter, the former series also diverges.