Proving $\sum_{1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $

Question

Show that $$\sum_{i=1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $$ where $ \lceil\cdot\rceil$ denotes the ceiling function.

My method: one way would be observe each part of the summation is telling numbers of form $2^\alpha k$ where $k$ is any odd number (from the list of numbers from $2$ to $2n$).

But I would like to know whether there is an algebraic proof, or maybe some bit of number theory that will help solve it.

Is this true? If $n=2$ the left hand is $\big \lfloor \log_2 \frac {4}{1}\big \rfloor + \big \lfloor \log_2 \frac {4}{3}\big \rfloor=2$ not $3$. — lulu, Sep 05 '22 at 11:39
Oh, you are using the ceiling function? That's very non-standard notation. You used the old notation for the floor function. I suggest editing to use the standard notation for the ceiling function, \lceil $\cdots$ \rceil — lulu, Sep 05 '22 at 11:44
Or you might be having a better way of getting the intuition of the above proof i gave to make the expression right — ProblemDestroyer, Sep 05 '22 at 11:44
I edited it to improve the notation, but I might have stomped on your notation. — Thomas Andrews, Sep 05 '22 at 11:46
@ProblemDestroyer FYI, using an Approach0 search, I found the more general Does $\sum_{i=1}^{k-1}\lceil \log_2\frac{N}{i}\rceil$ have a closed form?, and the somewhat similar How to show $n=1+\sum_{k=1}^{n}\left\lfloor{\log_2\frac{2n-1}{2k-1}}\right\rfloor$ for every natural number $n$.. — John Omielan, Sep 05 '22 at 17:38

score 2 · Answer 1 · answered Sep 05 '22 at 16:23

You can prove it by induction. The first observation is: $$\left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil = 1 + \left\lceil\log_{2}\frac{n}{2i-1}\right\rceil$$

So it suffices to prove $\sum_{i=1}^{n} \left\lceil\log_{2}\frac{n}{2i-1}\right\rceil = n-1$. The next observation is that the latter half of the sum is zero (when $2i - 1 > n$, we have $\frac{n}{2i-1} \in (\frac 12, 1)$). Formally:

$n=2k$:

$$\sum_{i=1}^{2k} \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = \sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil + 0,$$

which by induction hypothesis is $2k-1 = n - 1$.

$n=2k + 1$:

$$\sum_{i=1}^{2k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = \sum_{i=1}^{k + 1} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil + 0 = 1 + \sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil.$$

We want to show that $\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = (n - 1) - 1 = 2k$. By the induction hypothesis, we know that $\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = 2k - 1$. Let's compare them term by term, i.e. $\left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil$ vs $\left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil$.

Let's fix $i$, and let $\ell$ be the smallest integer such that $2^\ell \ge \frac{2k}{2i-1}$. When is it not the case that $2^\ell \ge \frac{2k + 1}{2i - 1}$? It happens iff $\frac{2k}{2i - 1}$ is a power of two. This happens for exactly on term: for $i$ such that $2k = (2i - 1) \cdot 2^t$ for some $t$. Hence, $$\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = 1 +\sum_{i=1}^k \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = 2k$$

Proving $\sum_{1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $

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