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.

Let $n\in\mathbb{Z}^{+}$ and $\displaystyle S_n = \sum_{k=1}^{n}\frac{n-k+1}{k}$. Finding $\Theta(S_n)$

PS: I found $\mathcal{O}(S_n) = n^2$. Thus, having $(n-k+1)/k = (n+1)/k -1 \leq n$.

$\rightarrow S_n = \sum_{k = 1} ^ {n}n = n^2$. But I cant find $\mathcal{\Omega}(S_n)$, so I cant also find $\mathcal{\Theta}(S_n)$.

share|improve this question
3  
The $i$ should be a $k$, right? Try writing it as $(n+1) \sum 1/k$ minus $\sum k/k = \sum 1$. You can deal with those sums separately... –  mt_ Dec 28 '11 at 16:39
    
@mt_: I'm sorry. I typed not correct. I editted. Thanks! –  qwerty89 Dec 28 '11 at 16:46
1  
@mt_ : perhaps you should make it an answer, asymptotics for $H_n$ are well-known, are they? –  Patrick Da Silva Dec 28 '11 at 16:47
    
I found $\mathcal{O}(S_n) = n^2$. Thus, having $(n-k+1)/k = (n+1)/k -1 \leq n$. ==> $S_n = \sum_{k = 1} ^ {n}n = n^2$. But I cant find $\mathcal{\Omega}(S_n)$, so I cant also find $\mathcal{\Theta}(S_n)$ –  qwerty89 Dec 28 '11 at 16:50
2  
qwerty89: Did you read the comments above? They suggest to first get/use a simple asymptotics of $H_n=\sum\limits_{k=1}^n\frac1k$. Did you do that? –  Did Dec 28 '11 at 17:01

1 Answer 1

up vote 4 down vote accepted

Let's recall what $\Theta$ means: "f is $\Theta(g)$" means that there are constants C,D such that for large enough $n$, $C g(n) \leq f(n) \leq Dg(n)$.

Your sum $S_n$ splits into $A_n-B_n$ where $A_n=(n+1)\sum_{k=1}^n 1/k $ and $B_n=\sum_{k=1}^n 1$.

$B_n=n$, so no problem with the asymptotics there: $B_n$ is $\Theta(n)$. What if I told you $\sum_{k=1}^n 1/k = \log(n) +\gamma_n$ where $(\gamma_n)$ is a convergent sequence. Could you find the asymptotics of $A_n$ then?

share|improve this answer
1  
Nice answer. You might consider replacing thrice $\backslash$sum by $\backslash$sum$\backslash$limits. –  Did Dec 28 '11 at 17:51
    
Well, as Concrete Mathematics points out, $H_n = \ln n + \gamma + (2n)^{-1} - (12n^2)^{-1} + (120n^4)^{-1} + O(n^{-6})$, where $H_n = \sum_{k=1}^n k^{-1}$. –  Frank Science Jun 8 '12 at 11:14

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.