Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Trying to find a closed form expression for the following sum, or an asymptotic expression in terms of well known functions (like the Gamma function, for instance).

Let $m,n$ be positive integers such that $2 \leq m < n$. Estimate the sum $$ \sum_{j=1}^m (-1)^j \binom m j \frac{\log(n-j)}{j} $$

where $\log$ stands for the natural logarithm. Thanks for any help.

share|improve this question
    
What you've written looks like an assignment question, though I could be wrong. –  David Roberts Oct 18 '12 at 6:04
1  
Similar problem posted to m.se with identical title a few days ago: math.stackexchange.com/questions/211449/… –  Gerry Myerson Oct 18 '12 at 6:34
    
I posted a similar problem on stackexchange, but that was a different problem and easier in that I have an answer in that case. The problem on stackexchange posted a few days ago was this: find a closed form expression for: $\sum_{j=1}^m (-1)^j \binom m j \log(1 - j/n)$. Notice the missing $1/j$ term in the sum. The answer is an asmptotic expression involving the $\Gamma$ function. –  krishnan.shankar Oct 18 '12 at 14:13
add comment

2 Answers

As $n \to \infty$ for fixed $m$, $$\log(n-j) = \log(n) + \log(1-j/n) = \log(n) - j/n + O(1/n^2)$$ Since $\sum_{j=1}^m (-1)^j {m \choose j} \frac{1}{j} = -\Psi(m+1)-\gamma$ and $\sum_{j=1}^m (-1)^j {m \choose j} = -1$, your sum is $-(\Psi(m+1)+\gamma) \log(n) + \dfrac{1}{n} + O\left( \dfrac{1}{n^2} \right) $.

share|improve this answer
1  
Yes, but fixed $m$ is the easiest case and Krishnan didn't specify that $m$ is fixed. It is a lot more challenging if $m,n$ both go to infinity. Try $n=m+1$. –  Brendan McKay Oct 18 '12 at 8:55
add comment

On of methods for finding an estimating or finding bound for a finite sum is using the following formula

$\sum_{n=a}^b f(n) \sim \int_a^b f(x)\,dx + \frac{f(a) + f(b)}{2} + \sum_{k=1}^\infty \,\frac{B_{2k}}{(2k)!}\left(f^{(2k - 1)'}(b) - f^{(2k - 1)'}(a)\right)$ and by taking $f(j)= (-1)^j\binom {m}{j}\frac{log(n-j)}{j}$ you can estimate right hand side of this formula, which are faster than of left hand side. In fact by computing some first terms of this infinite sum, we can obtain a good estimation for left hand side. Also here, $B_k$ are Bernoulli numbers.

Moreover, the sharp bounds of Bernoulli numbers has been computed ,(see here ) So you can also try to find a bound for your sum.

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.