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I am asking how to do the Limit comparison test for this: $$\lim {An\over Bn} =L$$ You choose $B_n$ yourselve, but how do you choose it?

Example:

$$A_n = \frac{3n^2 + 5n + 1}{\sqrt{(n^5 + 5 )}}$$

$$B_n = {3\over \sqrt{n}}$$

$$\large \lim {\frac{3n^2 + 5n + 1}{\sqrt{(n^5 + 5 )}}\over {3\over \sqrt{n}}} = 1$$

It can be concluded that $\lim A_n$ diverges.

Why? How to get $B_n$

and how to do the limit comparison test for this?

An example of another question: $$An={(2n+200)\over (e^\frac n3)-20}$$ An converges or diverges?

problematic
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3 Answers3

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Look at $A_n$. On the numerator the dominating term in $3n^3$, and on the denominator its $\sqrt{n^5}=n^2 \sqrt{n}$. Dividing the two gives $$\frac{3n^3}{n^2 \sqrt{n}}=3 \sqrt{n}. $$

user1337
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In such excercises, you shall take Bn to be a series that is "simpler" than what you have, but of same "order". To do so, you shouldl, roughly, spot the strongest parts of the numerator and the denominator, and divide them.

Mike
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  • strongest means highest power? so it is 3n^3 / n^5/2? =3n^1/2,but the answer is 3n^-1/2? what about the other example? how to solve it? – problematic Sep 14 '14 at 11:18
  • Strongest means the one that "beats" any other one, that is - it is bigger than any one else for sufficiently large n. In some cases it means indeed the highest power, but not always - you can have there e^n, n^n and so on, so it may be more complicated. – Mike Sep 14 '14 at 18:53
  • To solve your example, take Bn=sqrt(n), that is the division of n^3 and n^5/2. sqrt(n) diverges, thus An as well, by the comparison test. – Mike Sep 14 '14 at 18:56
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For $n$ sufficiently large, you can expand and get $$A_n = \frac{3n^3 + 5n + 1}{\sqrt{n^5 + 5 }}= \sqrt n \frac{3n^3 + 5n + 1}{\sqrt{n^6 + 5n }}\simeq \sqrt n \frac{3n^3}{\sqrt{n^6 }}=3 \sqrt n $$

More sophisticated would be a Taylor expansion $$A_n=3 \sqrt{n}+5 \left(\frac{1}{n}\right)^{3/2}+\left(\frac{1}{n}\right)^{5/2}+O\left(\left(\frac{1 }{n}\right)^{9/2}\right)$$

There is clearly a typo somewhere.

Added later fater the typo was fixed

If $$A_n = \frac{3n^2 + 5n + 1}{\sqrt{n^5 + 5 }}$$ it is obvious to ntoice that, for large values of $n$, $A_n \approx \frac{3n^2}{n^{5/2}}=\frac{3}{\sqrt n}$. THe expansion would be $$A_n=3 \sqrt{\frac{1}{n}}+5 \left(\frac{1}{n}\right)^{3/2}+\left(\frac{1}{n}\right)^{5/2}+O\left(\left(\frac{1 }{n}\right)^{7/2}\right)$$

Claude Leibovici
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