Problem A. 854. (May 2023)

A. 854. Prove that

\(\displaystyle \sum_{k=0}^n \frac{2^{2^k}\cdot 2^{k+1}}{2^{2^k}+3^{2^k}}<4 \)

holds for all positive integers \(\displaystyle n\).

Submitted by Béla Kovács, Szatmárnémeti

(7 pont)

Deadline expired on June 12, 2023.

First observe that

\(\displaystyle \dfrac{2^{2^k}\cdot 2^{k+1}}{2^{2^k}+3^{2^k}}= \dfrac{2^{k+1}}{(3/2)^{2^k}-1}-\dfrac{2^{k+2}}{(3/2)^{2^{k+1}}-1}. \)

Applying this we get a telescopic sum:

\(\displaystyle \sum_{k=0}^\infty \dfrac{2^{2^k}\cdot 2^{k+1}}{2^{2^k}+3^{2^k}} =\sum_{k=0}^\infty \Bigg(\dfrac{2^{k+1}}{(3/2)^{2^k}-1}-\dfrac{2^{k+2}}{(3/2)^{2^{k+1}}-1}\Bigg) =4, \)

and this proves the claim.

How is it possible to come up with this solution?

First, we want to get a telescopic sum, so we are looking for sequences \(\displaystyle a_0, a_1, \ldots\) and \(\displaystyle b_0, b_1, \ldots\) satisfying

\(\displaystyle \dfrac{2^{2^k}\cdot 2^{k+1}}{2^{2^k}+3^{2^k}}=\dfrac{a_k}{b_k}-\dfrac{a_{k+1}}{b_{k+1}}. \)

Let's rewrite the original algebraic expression:

\(\displaystyle \dfrac{2^{2^k}\cdot 2^{k+1}}{2^{2^k}+3^{2^k}}=\dfrac{2^{k+1}}{(3/2)^{2^k}+1}. \)

Now observe that

\(\displaystyle \left((3/2)^{2^k}+1\right)\left((3/2)^{2^k}-1\right)=(3/2)^{2^{k+1}}-1, \)

and this give the idea to choose \(\displaystyle b_k=(3/2)^{2^k}-1\). Substituting this into our equation we get

\(\displaystyle \dfrac{2^{k+1}}{(3/2)^{2^k}+1}=\dfrac{a_k}{(3/2)^{2^k}-1}-\dfrac{a_{k+1}}{(3/2)^{2^{k+1}}-1}=\frac{\big((3/2)^{2^k}+1\big)a_k-a_{k+1}}{\left((3/2)^{2^k}+1\right)\left((3/2)^{2^k}-1\right)}. \)

Let's multiply with the denominator of the right hand side:

\(\displaystyle 2^{k+1}\left((3/2)^{2^k}-1\right)=\left((3/2)^{2^k}+1\right)a_k-a_{k+1}. \)

Now it's easy to see that \(\displaystyle a_k=2^{k+1}\) will work.

Statistics:

6 students sent a solution.
7 points:	Bodor Mátyás, Diaconescu Tashi, Fleischman Illés, Szakács Ábel, Varga Boldizsár, Wiener Anna.

Problems in Mathematics of KöMaL, May 2023