I asked a (seed) question about sums of Landau terms before, trying to gauge the dangers of abusing asymptotics notation in arithmetics, with mixed success.

Now, over here our recurrence guru JeffE does essentially this:

$\qquad \displaystyle \sum_{i=1}^n \Theta\left(\frac{1}{i}\right) = \Theta(H_n)$

While the end result is correct, I think this is wrong. Why? If we add in all the existence of constants implied (only the upper bound), we have

$\qquad \displaystyle \sum_{i=1}^n c_i \cdot \frac{1}{i} \leq c \cdot H_n$.

Now how do we compute $c$ from $c_1, \dots, c_n$? The answer is, I believe, that we can not: $c$ has to bound for all $n$ but we get more $c_i$ as $n$ grows. We don't know anything about them; $c_i$ may very well depend on $i$, so we can not assume a bound: a finite $c$ may not exist.

In addition, there is this subtle issue of which variable goes to infinity on the left-hand side -- $i$ or $n$? Both? If $n$ (for the sake of compatibility), what is the meaning of $\Theta(1/i)$, knowing that $1 \leq i \leq n$? Does it not only mean $\Theta(1)$? If so, we can't bound the sum better than $\Theta(n)$.

So, where does that leave us? It it a blatant mistake? A subtle one? Or is it just the usual abuse of notation and we should not look at $=$ signs like this one out of context? Can we formulate a (rigorously) correct rule to evalutate (certain) sums of Landau terms?

I think that the main question is: what is $i$? If we consider it constant (as it is inside the scope of the sum) we can easily build counterexamples. If it is not constant, I have no idea how to read it.

  • 2
    $\begingroup$ This question on math.SE is a good read about arithmetics with Landau terms in general. $\endgroup$
    – Raphael
    Commented Jul 18, 2012 at 15:36
  • 4
    $\begingroup$ From the link you gave, the equality can be seen to be either a subset relation or an "is in" relation (i.e. $\in$). For $\Theta$ you're just saying it's bounded above and below by a constant. Why not choose $c = \min(c_1, c_2, \cdots, c_n)$ and $C = \max(c_1, c_2, \cdots, c_n)$? $\endgroup$
    – user834
    Commented Jul 18, 2012 at 17:33
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    $\begingroup$ Hang on there, Bucky. I didn't write any summation with a Theta in it. I wrote a recurrence with a Theta in it. Do you really interpret the recurrence "$t(n) = \Theta(1/n) + t(n-1)$" as something other than "There is a function $f \in \Theta_{x\to \infty}(x\mapsto 1/x)$ such that $t(n) = f(n) + t(n-1)$"? $\endgroup$
    – JeffE
    Commented Jul 19, 2012 at 16:06
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    $\begingroup$ @Raphael No, the recurrence is not mathematically the same as the sum, for precisely the reason you describe! The recurrence has exactly one Theta term in it, which unambiguously refers to a single function. $\endgroup$
    – JeffE
    Commented Jul 20, 2012 at 13:41
  • 2
    $\begingroup$ That's not very intuitive — I strongly disagree, but I think it's a matter of taste and experience. $\endgroup$
    – JeffE
    Commented Jul 20, 2012 at 17:01

3 Answers 3


Looks right to me in the following convention:

$S_n = \sum_{k=1}^{n} \Theta(1/k)$ is convenient notation for

There is an $f(x) \in \Theta(1/x)$ (as $x \to \infty$) such that

$S_n = \sum_{k=1}^{n} f(k)$.

Thus the $c_i$ (or with the notation in this answer $c_k$) you get, are not really dependent on $k$.

Under this interpretation, it is indeed true that $S_n = \Theta(H_n)$.

In fact, in Jeff's answer, he shows that $T(k+1) = f(k) + T(k)$ where $f \in \Theta(1/k)$, so it is consistent with the above interpretation.

The confusion seems to be arising from mentally "unrolling" the $\sum$ and presuming different functions for each occurrence of $\Theta$...

  • $\begingroup$ Jup, but every $\Theta$ can have its own function, and constant. So this convention does only work with context, that is if we know that the Landau terms stem from a somewhat "uniform" (in $k$ and $n$) definition of the summands. $\endgroup$
    – Raphael
    Commented Jul 19, 2012 at 11:40
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    $\begingroup$ @Raphael: It seems meaningless to unroll and then allow different $f_i$: the constants will then depend on the variable! and it becomes an incorrect usage of $\Theta$, assuming the $\Theta$ variable is $i$ (or $k$ in above answer). Even if we assume the variable is $n$, it is still looks meaningless to me. $\endgroup$
    – Aryabhata
    Commented Jul 19, 2012 at 15:06
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    $\begingroup$ In principle, every $\Theta$ can have its own constant, but in the particular context you describe, it's clear that every $\Theta$ does not have its own constant. $\endgroup$
    – JeffE
    Commented Jul 19, 2012 at 15:53
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    $\begingroup$ @JeffE: Right. We can have multiple $\Theta$ with their own constants, as long as the constants are really constant :-) $\endgroup$
    – Aryabhata
    Commented Jul 19, 2012 at 15:58
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    $\begingroup$ @JeffE So why don't you just write what you mean but prefer something ambiguous/wrong? Note that my updated answer now proposes a way to do so. I'd appreciate comments on that; downvotes with no reason don't help me understand why people seem to reject my point. $\endgroup$
    – Raphael
    Commented Jul 20, 2012 at 12:10

I think I nailed the problem down. In essence: using Landau terms decouples the variable of the summand function from the sum's running variable. We still (want to) read them as identical, though, therefore the confusion.

To develop it formally, what does

$\qquad \displaystyle S_n \in \sum_{i=1}^n \Theta(f(i)) \qquad \qquad\qquad (1)$

really mean? Now I assume that these $\Theta$ let $i$ -- not $n$ -- to infinity; if we let $n \to \infty$, every such sum evaluates to $\Theta(n)$ (if the summands are independent of $n$ and therefore constant) which is clearly wrong. Here is a first giveaway that we to crude things: $i$ is bound (and constant) inside the sum, but we still let it go to infinity?

Translating $(1)$ (for the upper bound, the lower bound works similarly), we get

$\qquad \displaystyle \exists f_1, \dots, f_n \in \Theta(f).\ S_n \leq \sum_{i=1}^n f_i(i)$

Now it is clear that the sum-$i$ and parameter-$i$ are decoupled: we can easily define the $f_i$ so that they use $i$ as a constant. In the example from the question, we can define $f_i(j) = i \cdot \frac{1}{j} \in \Theta(1/j)$ and have

$\qquad \displaystyle \sum_{i=0}^n f_i(i) "=" \sum_{i=0}^n \Theta(1/j) = \sum_{i=0}^n \Theta(1/i)$

but the original sum clearly does not evaluate to something in $\Theta(H_n) = \Theta(\log n)$. Now exchanging $j$ for $i$ -- which is only a renaming -- in the $\Theta$ may feel strange because $i$ is not independent of $n$ resp. the sum, but if we object to that now, we should never have used $i$ inside the $\Theta$ in the first place (as that holds the same strangeness).

Note that we did not even exploit that the $f_i$ may also depend on $n$.

To conclude, the proposed identity is bogus. We can, of course, agree on conventions on how to read such sums as abbreviation of rigorous calculation. However, such conventions will be incompatible with the definition of Landau terms (together with the normal abuse of them), impossible to understand correctly without context and misleading (for beginners) at least -- but that is ultimately a matter of taste (and ruthlessness?).

It occurred to me that we can also write exactly what we mean and still make use of the convenience of Landau terms. We know that all summands come from one common function, implying that the asymptotic bounds use the same constants. This is lost when we put the $\Theta$ into the sum. So let us not put it in there and write

$\qquad \displaystyle \sum_{i=1}^n \frac{2i - 1}{i(i+1)} \in \Theta\left(\sum_{i=1}^n \frac{1}{i}\right) = \Theta(H_n)$

instead. Putting the $\Theta$ outside of the sum results in

  • a mathematically correct statement and
  • a simple term inside the $\Theta$ we can easily deal with (which is what we want here, right?).

So it seems to me that this is both a correct and a useful way of writing the matter down, and should therefore be preferred over using Landau symbols inside the sum when we mean them outside of it.

  • $\begingroup$ Consider $\sum_i^n i$. I can define $f_i(n)=i$ (using $i$ as a constant), therefore $\sum_i^n i = \sum_i^n O(1) = O(n)$ by your reasoning, right? But this sum is $O(n^2)$. $\endgroup$
    – Xodarap
    Commented Jul 21, 2012 at 19:03
  • $\begingroup$ @Xodarap: By my reasoning, collapsing the sum like this does not work, because coupling the inner $\Theta$s (which are not coupled to $i$ nor $n$) to $n$ does change the meaning. $\endgroup$
    – Raphael
    Commented Jul 21, 2012 at 19:24
  • $\begingroup$ I'm not coupling them to $n$, I'm just using the fact that $\sum_i^n k = nk$. (And I suppose also the fact that $nO(f)=O(nf)$.) $\endgroup$
    – Xodarap
    Commented Jul 22, 2012 at 1:53
  • $\begingroup$ @Xodarap: But you don't have one $f$, but one $f_i$ per summand. If the underlying functions $f_i$ use $i$ (as a constant factor), you have to expand that, and the sum ends up being correct. So, clearly, by my reasoning the summing rule you propose does not work as you write. $\endgroup$
    – Raphael
    Commented Jul 22, 2012 at 8:51
  • $\begingroup$ If I have a sequence $5,1,3,2,\dots$, each of these are $O(1)$ (provided they don't increase as the series progresses). Would you say that adding $n$ of them will generate a sum $O(n)$? What's the difference if instead of being constants I describe them as constant functions $f_1(x)=5,f_2(x)=1,\dots$? $\endgroup$
    – Xodarap
    Commented Jul 22, 2012 at 14:03

If each $c_i$ is a constant, then there is some $c_{max}$ such that $\forall c_i: c_i\leq c_{max}$. So clearly $$\sum c_i f(i) \leq \sum c_{max} f(i) = c_{max} \sum f(i) = O\left(\sum f(i)\right)$$ Same idea for little o.

I think the problem here is that $1/i\not=\Theta(1)$. It's $o(1/n)$ (since there is no $\epsilon$ such that $\forall i: 1/i>\epsilon$), so the overall sum will be $no(1/n)=o(1)$. And each term is $O(1)$, meaning the overall sum is $O(n)$. So no tight bounds can be found from this method.

I think your questions are:

  1. Is bounding $\sum_i^n f(i)$ by doing the little o of each term and the big o of each term then multiplying by $n$ acceptable? (Answer: Yes)
  2. Is there a better method? (Answer: Not that I know of.)

Hopefully someone else can answer #2 more clearly.

EDIT: Looking over your question again, I think you are asking

$\sum_i^n \Theta(f(n)) = \Theta(nf(n))$ ?

To which the answer is yes. In this case though, each term is not $\Theta$ of anything, so that approach falls apart.

EDIT 2: You say "consider $c_i=i$, then there is no $c_{max}$". Unequivocally true. If you say that $c_i$ is a non-constant function of $i$, then it is, by definition, non-constant.

Note that if you define it this way, then $c_i i$ is not $\Theta(i)$, it's $\Theta(i^2)$. Indeed, if you define "constant" to mean "any function of $i$", then any two functions of $i$ differ by a "constant"!

Perhaps this is an easier way to think of it: we have the sequence $1,\frac{1}{2},\dots,\frac{1}{n}$. What's the smallest term in this sequence? Well, it will depend on $n$. So we can't consider the terms as constant.

(Computer scientists are often more familiar with big-O, so it might be more intuitive to ask if $1,\dots,n$ has a constant largest term.)

To provide your proof: let $f(i_{min})$ be the smallest value of $f(i)$ in the range $1,\dots,n$. Then $$\sum_i^n f(i) \geq \sum_i^n f(i_{min}) = n f(i_{min}) = n o(f(n))$$

An analogous proof can be made for the upper bound.

Lastly, you write that $H_n = o(n)$ and as proof give that $H_n = \Theta(\log n)$. This is in fact a counter-proof: if $H_n$ is "bigger" than $n$, then it can't be "smaller" than $\log n$, which is what's required for it to be $\Theta(\log n)$. So it can't be $o(n)$.

  • $\begingroup$ 1) "..then there is some $c_{max}$ such that..." -- no, there is not. Consider $(c_i)_{i \in \mathbb{N}}$ with $c_i = i$. 2) "I don't think $H_n=o(n)$" -- $H_n \in \Theta(\ln n)$ 3) $1/i\not=\Theta(1)$. It's $o(1/n)$ -- That is wrong. As $1/i \geq 1/n$, $1/i \in \Omega(1/n)$. 4) "(Answer: Yes)" -- as long as I don't see a formal proof of that fact, I don't believe it. Besides, "multiplying by $n$" is not what happened in the exhibited case. $\endgroup$
    – Raphael
    Commented Jul 18, 2012 at 21:32
  • $\begingroup$ I think you are missing the point. Your proof does not work because we may not have the same $f$ in every summand, and not even the same for the same summand but different $n$. I think I nailed it down; I will compose an answer shortly. $\endgroup$
    – Raphael
    Commented Jul 19, 2012 at 11:38
  • $\begingroup$ I still don't understand what you're saying, so I'm glad you figured it out :-) $\endgroup$
    – Xodarap
    Commented Jul 19, 2012 at 17:50

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