2
$\begingroup$

Let use the Pascal Triangle to create a sequence. We sum from the first term on the left to the term in the middle enter image description here

and then we take the inverse root times two. We get then that the terms are

\begin{eqnarray} &&a_0 := 2 \\ &&a_1 = 2 \frac{1}{1^{1/1}} = 2\\ &&a_2 = 2 \frac{1}{(2+1)^{1/2}} = 1.154700\dots\\ &&a_3 = 2 \frac{1}{(3+1)^{1/3}} = 1.25992\dots\\ &&a_4 = 2 \frac{1}{(6+4+1)^{1/4}} = 1.098200\dots\\ &&a_5 = 2 \frac{1}{(10+5+1)^{1/5}} = 1.14869\dots\\ &&a_6 = 2 \frac{1}{(20+15+6+1)^{1/6}} = 1.0727\dots\\ &&a_7 = 2 \frac{1}{(35+21+7+1)^{1/7}} = 1.10408\dots\\ \end{eqnarray}

The first line of the triangle creates the $a_0$ the second creates the $a_1$ and so on. Then I have my conjecture:

This sequence converges to 1

But I couldn't prove it or disprove it. What I did was tried to relate to the Binomial expansion: Let $0\leq x \leq 1$ so we have that

$$(1+x)^2 = 1 + 2x^2 + x^4 \geq x^2 + 2x^2 = 3x^2$$

So we have that

$$\frac{x}{x+1} \leq \frac{1}{\sqrt{3}}$$

And I tried to translate this to some limit using this $x$. There is a way to solve this problem like this? This series converges?

$\endgroup$
2
$\begingroup$

The sum of the elements in "half a row" of Pascal's triangle is either $2^{n-1}$ (if we are talking about the $n$-th row with $n$ being odd) or $2^{n-1}+\frac{1}{2}\binom{n}{n/2}$ if $n$ is even, due to $\binom{n}{k}=\binom{n}{n-k}$.
The central binomial coefficient $\binom{2n}{n}$ behaves in the following way $$ \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}} $$ hence you are essentially asking about $$ \lim_{n\to +\infty} 2\cdot\frac{1}{(2^{n-1}+E(n))^{1/n}} $$ that is clearly $1$.

$\endgroup$
  • $\begingroup$ @HazemOrabi: oh, many thanks, now fixed. $\endgroup$ – Jack D'Aurizio Mar 2 '17 at 1:50
  • $\begingroup$ What is $E(n)$? $\endgroup$ – Rafael Wagner Mar 2 '17 at 2:41
  • $\begingroup$ @RafaelWagner: an error term that is $o(2^{n-1})$. $\endgroup$ – Jack D'Aurizio Mar 2 '17 at 8:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.