Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to find a limit, or approximation for $\sum\limits_{k=1}^{n} (-1)^k {n \choose k} \log(a+bk)$ for, say, an $a,b\in (0,10)$. It is not so important what values $a$ and $b$ have. It would be already helpful for me to find a limit or an approximation for, say, $\sum\limits_{k=1}^{n} (-1)^k {n \choose k} \log(1+2k)$.

This sum arised in some analysis of stochastic processes and I have unfortunately almost not much knowledge in combinatorics and analysis to solve such equations. Thanks for any help.

share|improve this question
1  
Closely related: math.stackexchange.com/questions/64971/… –  David Speyer Mar 27 '12 at 15:51
    
Thanks, I will look through it. But in fact I know this limit (sorry should have mentioned this). Though I haven't yet tried to understand the proof because it seemed hard and I am not sure if it helps me with my equation. The serie in my question arises from a generalization where the easy case leads to the sum $\sum\limits_{k=1}^n \binom{n}{k}(-1)^k \log k$. –  Coopi Mar 27 '12 at 15:59
    
@Coopi: because your account is not registered, when you change computers or IP addresses the system sometimes "forgets" who you are. Please consider registering your account: this will guarantee you the ability to edit and comment on your own questions. –  Willie Wong Apr 4 '12 at 12:29
add comment

2 Answers 2

Set $r=a/b$. We are fixing $r$ once and for all, so constants in big $O$'s can depend on $r$.

We are dealing with $$\sum_{k=1}^{n} (-1)^k \binom{n}{k} \left( \log(1+k r^{-1}) + \log a \right) = - \log a + \sum_{k=0}^n (-1)^k \binom{n}{k} \left( \log(k+r) - \log r \right)$$ We can get away with changing the lower bound of the sum from $1$ to $0$ because that term is $0$ anyway.

Following the approach from the previous question, $$\sum_{k=0}^n (-1)^k \binom{n}{k} \left( \log(k+r) - \log r \right)= \int_{r}^{r+1} \frac{(n-1)! dx}{x(x+1)(x+2) \cdots (x+n-1)}.$$ You should double check me on this, but I get that this time there is no extra $1/x$ term to keep track of. (Of course, if I got that wrong, just adjust the final answer by $\int_r^{r+1} dx/x = \log(r+1) - \log r$.) WARNING I think I might have missed a $(-1)^{n-1}$ here. At least, when I tried to do $n=2$ by hand, I got $-\int_r^{r+1} dx/x(x+1)$, not $\int_r^{r+1} dx/x(x+1)$. I'm not going to fix this because of the likelihood that I'd introduce new errors; just leaving a warning for you to check it carefully.

As before, this is $$\int_{r}^{r+1} e^{-x \log n} (1+O(1/n)) \Gamma(x) dx = n^{-r}(1+O(1/n))\int_0^1 e^{-y \log n} \Gamma(r + y) dy.$$

Now for a standard trick. That exponential means that almost all the contribution to the integral will come from $y$ near $0$. Make the change of variables $y \log n =z$, and then the significant contribution will be spread out over the entire $z$ range.

In other words: $$\int_0^1 e^{-y \log n} \Gamma(1+r + y) dy = \frac{1}{\log n} \int_0^{\log n} e^{-z} \Gamma(r+z/\log n) dz.$$ And $$ \int_0^{\log n} e^{-z} \Gamma(r+z/\log n) dz = \int_0^{\log n} e^{-z} \left( \Gamma(r) + O(z/\log n) \right) dz = \Gamma(r) \int_0^{\log n} e^{-z} dz + O \left( \frac{1}{\log n} \int_0^{\infty} z e^{-z} \right)=$$ $$\Gamma(r)(1 - O(1/n)) + O(1/\log n) = \Gamma(r)(1+O(1/\log n)).$$

Putting back the terms we dropped along the way, I get $$\sum_{k=1}^n (-1)^k \binom{n}{k} \log(a+bk) = - \log a + \frac{\Gamma(r)}{n^r \log n} (1+O(1/\log n)).$$ If you started your sum at $k=0$ instead of $k=1$, that $- \log a$ term would nicely go away, which makes me wonder whether that is what you intended.

If you need more terms, you should be able to get an asymtotic series in powers of $1/\log n$ by approximating $\Gamma(r+1+z/\log n)$ by more terms of its Taylor series.

share|improve this answer
    
Thanks a lot. I will try to understand it. It seems somehow easier than the proof in the previous question (just $\log(k)$), right? Yes, the $-\log(a)$ in your last line cancels out, or same I could from the beginning have started the sum at 0. –  Coopi Mar 28 '12 at 10:26
    
@Coopi It looks easier for three reasons. One is that the domain of integration no longer contains $0$. The second is that I only got the leading term (analogous to $\log \log n$), not the first several. The third is that I didn't retype all the steps which were the same in both solutions. –  David Speyer Mar 28 '12 at 12:24
    
I don't know how to make a comment to the previous answer. But I think you forgot the $\frac{1}{x}$ in $\int_{r}^{r+1} e^{-x \log n} (1+O(1/n)) \Gamma(x+1) dx$, or you should use $\int_{r}^{r+1} e^{-x \log n} (1+O(1/n)) \Gamma(x) dx$ instead. So the answer should be (starting the sum from $0$): $\sum_{k=0}^n (-1)^k \binom{n}{k} \log(a+bk) =\frac{\Gamma(r)}{n^r \log n} (1+O(1/\log n)$? Is it ok to write $\sum_{k=0}^n (-1)^k \binom{n}{k} \log(a+bk) \approx \frac{\Gamma(r)}{n^r \log n}$, and so actually $\lim_{n \to \infty} \sum_{k=0}^n (-1)^k \binom{n}{k} \log(a+bk) = 0$ for $r$ fixed? –  Coopi Apr 4 '12 at 5:58
    
@David: Coopi added an answer with some follow-up questions. –  t.b. Apr 4 '12 at 10:00
    
You're right, I'll edit that $1/x$ factor in a bit. (I also think I missed a $(-1)^n$ somewhere.) I'm not sure why you have $(\log n)^2$ in the denominator, it should just be $\log n$. But, up to fixing that, I think I agree with everything you wrote. –  David Speyer Apr 4 '12 at 10:44
show 1 more comment

Recalling the identity of Stirling numbers of the second kind in terms of the binomial,

$$\left\{\begin{matrix} n \\ m \end{matrix}\right\} = \frac{1}{m!}\sum_{j=0}^m (-1)^{m-j}{m \choose j } j^n\,. $$

Writing the sum as $$ \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a) + \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(1+\frac{b k}{a})$$

$$ -\ln(a) + \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(1+\frac{b k}{a}) \,.$$

Expanding the $\ln(1+\frac{b k}{a})$ in terms of its Taylor series and interchanging the order of summation gives $$ -n!\,\sum_{m=1}^{\infty}{\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \left(\frac{1}{n!}\sum\limits_{k=1}^{n} (-1)^{m-k} {n \choose k} k^m \right) = -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} $$

$$ = -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,,$$

$$ \Rightarrow \sum\limits_{k=1}^{n} (-1)^k {n \choose k} \ln(a+bk) = -\ln(a) -n!\,\sum_{m=1}^{\infty}\left\{\begin{matrix} m \\ n \end{matrix}\right\} {\left(\frac{b}{a}\right)}^{m}\frac{1}{m} \,. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.