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# Multiple tests to check convergence Watch

1. Can you do something like use the Ratio Test to check convergence, and then when you simplify the a(n+1) / a(n) equation, use the alternating series test or something to check if that the new "equation" you have converges?
2. Hmm.. the short answer is no, I don't think so.

(the long answer is what do you mean, and can you give an example of what you're trying to do?)
3. (Original post by claret_n_blue)
Can you do something like use the Ratio Test to check convergence, and then when you simplify the a(n+1) / a(n) equation, use the alternating series test or something to check if that the new "equation" you have converges?
When you use the ratio test, you are left with checking the limit of a sequence. The alternating series test only applies to series, so it can't be applied.
4. (Original post by IrrationalNumber)
When you use the ratio test, you are left with checking the limit of a sequence. The alternating series test only applies to series, so it can't be applied.
Oh ok, and then I let n tend to infinity and see if it is bigger than, less than or equal to 1?

1 more question, if you have [a^(n +1) + c] / [b^n + d] where a < b and a,b,c and d are all positive integers, then will this tend to infinity because the b^n tends to 0, or will it tend to 1 because you will effectively have infinity / infinity?
5. (Original post by claret_n_blue)

1 more question, if you have [a^(n +1) + c] / [b^n + d] where a < b and a,b,c and d are all positive integers, then will this tend to infinity because the b^n tends to 0
b^n doesn't tend towards 0. Infact, it necessarily diverges to infinity (b^n that is) because b>a>=1.

or will it tend to 1 because you will effectively have infinity / infinity?
Never, ever, try and reason like that. For example, suppose we had the sequence n/n^2. The numerator tends towards infinity, the denominator tends towards infinity. The sequence itself tends towards 0 though: n/n^2= 1/n.

Instead, find out which term goes fastest (hint: b>a so b^n>a^n). Divide by this term, and then try and solve the problem.
6. (Original post by IrrationalNumber)
b^n doesn't tend towards 0. Infact, it necessarily diverges to infinity (b^n that is) because b>a>=1.
Oh sorry, b^n doesn't tend to 0, 1/(b^n) tends to 0, right?

(Original post by IrrationalNumber)
Never, ever, try and reason like that. For example, suppose we had the sequence n/n^2. The numerator tends towards infinity, the denominator tends towards infinity. The sequence itself tends towards 0 though: n/n^2= 1/n.

Instead, find out which term goes fastest (hint: b>a so b^n>a^n). Divide by this term, and then try and solve the problem.
I have simplified it as much as I could, but it is in the form that I gave the general form to. Would I just ignore the constants as the numbers would get so big that they would have little impact?

I think I understand what to do from that hint, but I am assuming that b^n > a^m, where m = n+1, because that is just a number, and like the other constant, if we let n be very big, the 1 will play little part, correct?

The main problem I am having is that seeing the base of a^(n+1) is smaller than b^n, but the power of a^(n+1) is bigger than the power of b^n, so I'm not sure which is the more "dominant" one. What is more important, power or base? I would've thought that the b^n tends quicker than the a^(n+1) because the base is bigger, but as the power of a^(n+1) is bigger, I am not sure if that will now tend quicker.

If this was the exam, and I had to make a choice, I think I would go for b^n tends to infinity quicker than a^(n+1), and so the whole thing will tend to 0.
7. (Original post by claret_n_blue)
Oh sorry, b^n doesn't tend to 0, 1/(b^n) tends to 0, right?

I have simplified it as much as I could, but it is in the form that I gave the general form to. Would I just ignore the constants as the numbers would get so big that they would have little impact?

I think I understand what to do from that hint, but I am assuming that b^n > a^m, where m = n+1, because that is just a number, and like the other constant, if we let n be very big, the 1 will play little part, correct?

The main problem I am having is that seeing the base of a^(n+1) is smaller than b^n, but the power of a^(n+1) is bigger than the power of b^n, so I'm not sure which is the more "dominant" one. What is more important, power or base? I would've thought that the b^n tends quicker than the a^(n+1) because the base is bigger, but as the power of a^(n+1) is bigger, I am not sure if that will now tend quicker.

If this was the exam, and I had to make a choice, I think I would go for b^n tends to infinity quicker than a^(n+1), and so the whole thing will tend to 0.
Ok, so you guess b^n goes faster. Try dividing numerator and denominator by that term, and see what happens.
8. (Original post by IrrationalNumber)
Ok, so you guess b^n goes faster. Try dividing numerator and denominator by that term, and see what happens.
When you say "very big", can I let it be something like 100 or 1000, or is that too small?
9. (Original post by claret_n_blue)
When you say "very big", can I let it be something like 100 or 1000, or is that too small?
When did I say very big? You can't just substitute a value for n in, it may not be large enough (for example, log(n) diverges to infinity, but if you set n=10^50, you only get 50log(10)<50log(e^4)=200, which is not very large considering the size of n)
10. (Original post by IrrationalNumber)
When did I say very big? You can't just substitute a value for n in, it may not be large enough (for example, log(n) diverges to infinity, but if you set n=10^50, you only get 50log(10)<50log(e^4)=200, which is not very large considering the size of n)
Oh sorry, my mistake. You didn't.

Hang one, I think I have misunderstood what you said. When you say divide by b^n, do you mean do something like:

[ (a^n/b^n) + (c/b^n) ] / [ 1 + (d/b^n) ]

And then do each term separately?

I tried to ask help from my tutor as well but he just gave me a list of books and I can't get to the uni library as I am back home.
11. (Original post by IrrationalNumber)
b^n doesn't tend towards 0. Infact, it necessarily diverges to infinity (b^n that is) because b>a>=1.

Never, ever, try and reason like that. For example, suppose we had the sequence n/n^2. The numerator tends towards infinity, the denominator tends towards infinity. The sequence itself tends towards 0 though: n/n^2= 1/n.

Instead, find out which term goes fastest (hint: b>a so b^n>a^n). Divide by this term, and then try and solve the problem.
Or you could just times every by -12341209482340891723508457023457 23948572034952734032745923485723 97239045729345723045712938471293 471234123487139487123471034....

You'll get the answer in the end.....
12. (Original post by claret_n_blue)
Oh sorry, my mistake. You didn't.

Hang one, I think I have misunderstood what you said. When you say divide by b^n, do you mean do something like:

[ (a^n/b^n) + (c/b^n) ] / [ 1 + (d/b^n) ]

And then do each term separately?

I tried to ask help from my tutor as well but he just gave me a list of books and I can't get to the uni library as I am back home.
That's exactly what I mean, though it should be a^(n+1)/(b^n). So for example, c/b^n goes towards 0 as n tends towards infinity. Think about what the bottom line does and why, and then think about what the top line does and why, and try and find the answer using that.
13. (Original post by IrrationalNumber)
That's exactly what I mean, though it should be a^(n+1)/(b^n). So for example, c/b^n goes towards 0 as n tends towards infinity. Think about what the bottom line does and why, and then think about what the top line does and why, and try and find the answer using that.

Ohhhhhh, thank you SOO much. I was really confused when you first said to divide everything by b^n. This is VERY helpful. Thank you very very much
14. (Original post by claret_n_blue)
Ohhhhhh, thank you SOO much. I was really confused when you first said to divide everything by b^n. This is VERY helpful. Thank you very very much
No problem! Have a nice christmas, hope the analysis becomes easier

The general trick here is to divide by the dominant term if you think the sequence will tend towards 0. In this case, it looks like b^n, so we try dividing by that.

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Updated: December 25, 2010
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