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.

Estimate following sums as the functions of variable $n$:

a) $\displaystyle\sum_{i=1}^{n}e^i\ln i$

b) $\displaystyle\sum_{i=1}^{2n}(-1)^i\ln i$

c) $\displaystyle\sum_{i\ge 0}^{}\frac{\ln(n+i)}{e^i}$

d) $\displaystyle\sum_{i\ge 0}^{}\frac{(-1)^i}{\ln(n+i)}$

I prefer the easiest ways to do such things, so Euler-Maclaurin formula isn't in my style. Are these examples very hard? I want to learn how to do such things but unfortunately when it comes to solving I have no ideas.

Only in b) I thought I knew but the idea is probably useless: $$\sum_{i=1}^{2n}(-1)^i\ln i=\sum_{i=1}^{n}\ln(2i)-\ln(2i-1)$$ so let: $$f(x)=\ln(2x)-\ln(2x-1)$$ and it's easy to check that $f(x)$ non increasing for $x\ge 1$ then: $$\int_i^{i+1}f(x)\le f(i)\le\int_{i-1}^if(x)$$ it follows: $$\int_2^{n+1}f(x)\le\sum_{i=2}^n f(i)\le\int_1^{n}f(x)$$ and integral is easy to count: $$\int f(x)=\frac{1}{2}\left( 2x\ln(2x)-(2x-1)\ln(2x-1)-1 \right)+C$$ But is it worth anything? Can I write from this $\sum_{i=1}^n f(i)$ using asymptotics notations like big $O$ or $\Theta$?

I don't know how to approach the rest.

share|improve this question
    
The first sum is dominated by its largest term; try to show that it is in fact $\Theta(e^n \log n)$. –  Qiaochu Yuan Aug 24 '12 at 18:35
    
Why do you feel like Euler-Maclaurin isn't 'the easiest way' to solve problems like this? –  Steven Stadnicki Aug 24 '12 at 18:39
    
@QiaochuYuan, I'm not sure how to do that. –  ray Aug 24 '12 at 19:58
    
Perhaps you might want to actually compute a few values of the sum to get a sense of how the terms grow. –  Qiaochu Yuan Aug 24 '12 at 19:59
    
@StevenStadnicki, I am ashamed to admit, but this formula seems too hard for me. I have even problems with using it.. –  ray Aug 24 '12 at 20:00

2 Answers 2

up vote 2 down vote accepted

For case (b), introduce $x_k=\log(2k)-\log(2k-1)\sim\frac1{2k}$. Then, $s_n=\sum\limits_{k=1}^{2n}(-1)^k\log(k)=\sum\limits_{k=1}^nx_k$ is such that $$ s_n\sim\sum\limits_{k=1}^n\frac1{2k}\sim\frac12\log(n). $$ To be more precise, note that $x_k=\int\limits_{2k-1}^{2k}\frac{\mathrm dt}t$ hence $2x_k\leqslant\int\limits_{2k-2}^{2k}\frac{\mathrm dt}t$ for every $k\geqslant2$ and $2s_n\leqslant2x_1+\int\limits_2^{2n}\frac{\mathrm dt}t=2\log(2)+\log(2n)-\log(2)$, that is, $s_n\leqslant\frac12\log(n)+\log(2)$ for every $n\geqslant1$. Likewise, $2x_k\geqslant\int\limits_{2k-1}^{2k+1}\frac{\mathrm dt}t$ for every $k\geqslant1$ and $2s_n\geqslant\int\limits_1^{2n+1}\frac{\mathrm dt}t=\log(2n+1)$, that is, $s_n\geqslant\frac12\log(n)+\frac12\log(2)+\log(1+\frac1{2n})$ for every $n\geqslant1$. This yields the asymptotics $$ s_n=\frac12\log(n)+O(1). $$ Edit: Roughly the same techniques of comparison with integrals yield the three other cases, but it might be more profitable to understand fully case (b) before studying them. Anyway, unless I am mistaken, simple equivalents for cases (a), (c) and (d) are $$ \text{(a)}\ \frac{\mathrm e}{\mathrm e-1}\log(n)\,\mathrm e^n,\quad \text{(c)}\ \frac{\mathrm e}{\mathrm e-1}\log(n),\quad \text{(d)}\ \frac1{2\log n}. $$

share|improve this answer

For b, you can say the sum is $\ln \frac {(2n)!!}{(2n-1)!!}=\ln \frac {2^n(n!)^22^n}{(2n)!}$ and then use Stirling's approximation.

share|improve this answer

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.