Asymptotics analysis of this expression

Question

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.

Could someone help? many thanks!

Answer 1

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 $a = 1/2$ $$u_n \sim \log n - \frac{1}{4} \to \infty$$

Answer 2

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): $$f(n) \sim a_2 n^{2\alpha + 1} = -\frac{1}{4}n^{2\alpha + 1}\rightarrow -\infty.$$