# Limit sequences converges

1. Jan 12, 2006

### teng125

determine the following sequences converges 17n^54 + 1/n^2 +42 divide by n^55 + 75n^54....

pls help....

2. Jan 12, 2006

### Mindscrape

Looks like something to do on a graphing calculator. Basically you find the limit as n approaches infinity and negative infinity, and see if it converges on a number by pluging in larger and larger values of n. So, for example, that is our function, let's say f(n), and you find that f(-10)=.344 (I'm just making this up, the actual function could be different), f(-100)=.488, f(-1000)=.499998, then the function looks like it converges on .5.

3. Jan 12, 2006

### teng125

but i divide every part by n^54 and find the limit and i got 17/(n+75)
then it goes to zero.is it rite??

4. Jan 12, 2006

### TD

If I understand the problem correctly, it tends to zero indeed.

5. Jan 12, 2006

### HallsofIvy

Staff Emeritus
Before anyone can determine whether it converges or not you will have to tell us what the sequence really is. Just telling us four terms is not enough. What is the "rule" or general term in the sequence?

6. Jan 12, 2006

### teng125

no,the question is eactly same as the above

7. Jan 12, 2006

### TD

I'm assuming you mean the sequence with general term:

$$u_n = \frac{{17n^{54} + \frac{1} {{n^2 }} + 42}} {{n^{55} + 75n^{54} }}$$

8. Jan 12, 2006

### teng125

yaya......that's correct

9. Jan 12, 2006

### HallsofIvy

Staff Emeritus
Oh, I see. I thought what you had written was the series, depending on a parameter n. Divide both numerator and denominator by n55 and it should be easy to see that the sequence converges to 0. As a series or sum, you can "compare" it to 1/n and see that even though the series of terms converges to 0, the series diverges.