Limit theorems and determining convergence

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Homework Statement



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The Attempt at a Solution



I'm having some trouble getting my head around these 3 problems. Any ideas on how to approach them are welcome.
 
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Can you show us what you have tried?

You should be able to simplify the expressions in a) and c).

For b), take a look at the numerator and see what the values for the first few n are. Notice that n! is always an integer, so what can you determine about the possible values of sin(n! * pi /3) ? How does the (-1)^n affect those possible values? Can you determine a maximum possible value and a minimum possible value for the numerator? If so, what will happen to the value of the whole expression as n gets larger in the denominator?
 
The first few limits are straightforward. For part a), algebraic manipulation will get you to the correct answer (it does exist; I'll let you figure out what it is :). Try to think of how you can use the various powers of n. For part b), clearly, as n! gets larger, it contains a factor of 3 which cancels with the 3 in the denominator of the sine function. Thus, the sine function goes to zero and the sandwich theorem can be applied with ease.

As an edit, think about how you can simplify the third expression to get a limit that is far more intuitive to work with. After simplification, the limit will follow naturally.
 
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Thanks for the replies!

I'm having some trouble manipulating the denominator of a).

For b), I can see that whenever n>3 the sin term is 0. Since the (-1)^n-1 oscillates, does the expression diverge by unbounded oscillation?

For c), I can simplify it up to x^(3 - logx). After that I can't seem to go any further.
 
Oshada said:
Thanks for the replies!

I'm having some trouble manipulating the denominator of a).

For b), I can see that whenever n>3 the sin term is 0. Since the (-1)^n-1 oscillates, does the expression diverge by unbounded oscillation?

No, as the oscillation isn't unbounded. It clearly tends to a specific number. Try neglecting the sine term (as you can for n>3), and just look at (-1)^(n-1)/(n+1). As n increases, what does this get closer to?
 
Can it be... 0?!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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