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How to analyze the asymptotics of $\log(n-1)-\frac{1}{2}n^{\alpha+1}+\frac{1}{2}n\log(1+n^\alpha)$? $\alpha<0$, when $n$ goes to infinity. When does it goes to $-\infty$ when $n$ goes to $+\infty$?

I have no experience of dealing with this type. Should I expand them in Taylor series?

I tried so:

$$=\log(n-1)-\frac{n^{2\alpha+1}}{2}+\frac{n^{3\alpha+1}}{3}-\frac{n^{4\alpha+1}}{4}+O(n^{5\alpha+1})\\ \prec(n-1)-\frac{n^{2\alpha+1}}{2}+\frac{n^{3\alpha+1}}{3}-\frac{n^{4\alpha+1}}{4}+O(n^{5\alpha+1})$$

But then I still feel no where to put my hands on..

Numerics tell that it seems when around $a>-0.45$ the expression goes to negative infinity. But I don't know how to prove this.

enter image description here Could someone help? many thanks!

happyle
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  • Hints: * when $u$ is small, $\log(1+u)$ is roughly $u$; * so that makes the second and third terms about the same size with opposite signs—expect cancellation; * one way of quantifying that cancellation is indeed with Taylor series with remainder; try applying that to $g(u) = \frac12n\log(1+u)$. – Greg Martin Sep 08 '23 at 16:51
  • I think the idea is good, however I would not touch the log – julio_es_sui_glace Sep 08 '23 at 16:52

2 Answers2

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The limit is indeed tricky I'm not sure how precise the answer needs to be, so I'll give a nice asymptotic expansion.

Let us cut this sequence in two: $$u_n = \log(n-1) - \frac{1}{2}n^{\alpha +1} + \frac{1}{2}n \log(1 + n^\alpha) = \log(n-1) + v_n$$

Let us study $v_n$, since $n^\alpha \to 0$ when $n \to \infty$, we can use Taylor expansion:
$\begin{align*} v_n & = - \frac{1}{2}n^{\alpha +1} + \frac{1}{2}n \log(1 + n^\alpha) = - \frac{1}{2}n^{\alpha +1} + \frac{1}{2}n \left(n^\alpha + -\frac{n^{2\alpha}}{2} + \cdots + (-1)^{m+1}\frac{n^{m\alpha}}{m} + O(n^{(m+1)\alpha})\right)\\ &= - \frac{1}{2}n^{\alpha +1} + \frac{1}{2} n^{\alpha+1} - \frac{n^{2\alpha +1}}{4} + \cdots +(-1)^{m+1} \frac{n^{m\alpha + 1}}{2m} + O(n^{(m+1)\alpha +1})\\ &= -\frac{n^{2\alpha +1}}{4} + \cdots + (-1)^{m+1}\frac{n^{m\alpha + 1}}{2m} + O(n^{(m+1)\alpha +1}) \end{align*}$

On the other hand, $\log(n-1) = \log(n) + \log (1 - \frac{1}{n}) = \log(n) - n^{-1} - \cdots - \frac{1}{p}n^{-p} + O(n^{p+1})$

Then the placement depends on the value of $a$, if $a < -1$ then $$u_n = \log n - 1/n + o(1/n)$$ if $-1<a<0$ then $$u_n = \log n - \frac{n^{2\alpha +1}}{4} + o(n^{2\alpha +1})$$ if $a = -1$ then $$u_n = \log n - \frac{5}{4n} + o(1/n)$$

Otherwise you can use as asymptotic expansion $$u_n = \log(n-1) - \frac{n^{2\alpha +1}}{4} + \cdots + (-1)^{m+1} \frac{n^{m\alpha + 1}}{2m} + O(n^{(m+1)\alpha +1})$$

We can see that for $a< -1/2$, $u_n \sim \log n \to \infty$ and for $a > -1/2, u_n \sim \frac{n^{2\alpha +1}}{4} \to -\infty$, now for $a = 1/2$ $$u_n \sim \log n - \frac{1}{4} \to \infty$$

julio_es_sui_glace
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  • What is $a$? Is it $\alpha$? – Gary Sep 08 '23 at 22:28
  • Thank you so much! I have a follow-up question and and many thanks in advance - So I think for both the two series, the first term dominates, thus Can I write $-\frac{n^{2\alpha +1}}{4} + \cdots + (-1)^{m+1}\frac{n^{m\alpha + 1}}{2m} + O(n^{(m+1)\alpha +1})$ as $ O(-\frac{n^{2\alpha +1}}{4})$? and write $- n^{-1} - \cdots - \frac{1}{p}n^{-p} + O(n^{p+1})$ as $O(- n^{-1} )$? Then I just need to study $\log n+O(-\frac{n^{2\alpha +1}}{4})+O(- n^{-1} )$ which form looks simpler.. – happyle Sep 09 '23 at 10:02
  • You can choose $m=2$ for example, I juste wrote the complete Taylor expansion, but it was not needed! – julio_es_sui_glace Sep 09 '23 at 18:07
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If $\alpha < 0$, then $n^{\alpha}\rightarrow 0$ as $n\rightarrow\infty$, and so terms involving $n^\alpha$ can be expanded accordingly (i.e., in powers of $n^\alpha$). So $$ \frac{1}{2}n\log(1+n^{\alpha})\sim \frac{1}{2}n\left(n^\alpha-\frac{1}{2}n^{2\alpha}+\frac{1}{3}n^{3\alpha}-\ldots\right). $$ The first term of this expansion and the $-(1/2)n^{1+\alpha}$ term cancel each other out. Also, since $\log(n-1)=\log n + \log(1-1/n)$, that term should be expanded in powers of $1/n$ (which is also going to $0$).

Putting it together, you have a logarithmic term plus the sum of two asymptotic series: $$ f(n) \sim \log n + \sum_{k=2}^{\infty} a_k n^{1+k\alpha}+\sum_{m=1}^{\infty}\frac{b_m}{n^m}. $$ If $\alpha \le -1/2$, then the logarithmic term (which is positive) dominates: $$ f(n) \sim \log n \rightarrow \infty. $$ On the other hand, if $\alpha > -1/2$, then the leading-order behavior comes from the first term in the $a_k$ series (which is negative): $$f(n) \sim a_2 n^{2\alpha + 1} = -\frac{1}{4}n^{2\alpha + 1}\rightarrow -\infty.$$

mjqxxxx
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  • thank you! About "if $\alpha<-1/2$ then $log$ term dominates." - I thought about this, and here is my understanding - when $\alpha<-1/2$ then in the first summation, all terms $n^{1+k\alpha}<0$ thus the summation goes to zero. Moreover, the second summation goes to zero as well. Thus $\log n$ goes to $+\infty$ and two summations goes to $0$. Thus overall goes to $+\infty$. Is this what you meant explictly by "logarithmic term dominates"? Thank you again. – happyle Sep 09 '23 at 09:11
  • Yes, correct... at least, all terms $n^{1+k\alpha}$ are $o(\log n)$. (The summation doesn't need to go to zero; it just needs to become much smaller than $\log n$ as $n\rightarrow\infty$. In particular, if $\alpha=-1/2$, the leading term doesn't vanish.) – mjqxxxx Sep 09 '23 at 18:48