Can we determine if the limit sequence converges?

  • Thread starter Thread starter teng125
  • Start date Start date
  • Tags Tags
    Limit Sequences
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
8 replies · 2K views
teng125
Messages
416
Reaction score
0
determine the following sequences converges 17n^54 + 1/n^2 +42 divide by n^55 + 75n^54...

pls help...
 
Physics news on Phys.org
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.
 
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??
 
teng125 said:
determine the following sequences converges 17n^54 + 1/n^2 +42 divide by n^55 + 75n^54...
pls help...
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?
 
no,the question is eactly same as the above
 
I'm assuming you mean the sequence with general term:

[tex]u_n = \frac{{17n^{54} + \frac{1}<br /> {{n^2 }} + 42}}<br /> {{n^{55} + 75n^{54} }}[/tex]
 
yaya...that's correct
 
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.