Can we determine if the limit sequence converges?

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The discussion focuses on determining the convergence of the sequence defined by the expression (17n^54 + 1/n^2 + 42) / (n^55 + 75n^54). Participants suggest using a graphing calculator to evaluate the limit as n approaches infinity. By dividing both the numerator and denominator by n^55, it becomes clear that the sequence converges to 0. There is also a mention of comparing the sequence to 1/n to illustrate that while the terms converge to 0, the series diverges. Overall, the consensus is that the sequence converges to 0.
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determine the following sequences converges 17n^54 + 1/n^2 +42 divide by n^55 + 75n^54...

pls help...
 
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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??
 
If I understand the problem correctly, it tends to zero indeed.
 
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:

u_n = \frac{{17n^{54} + \frac{1}<br /> {{n^2 }} + 42}}<br /> {{n^{55} + 75n^{54} }}
 
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.
 
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