Can You Prove This Reciprocal Sequence Sum is Less Than 4?

[anemone]

Here is this week's POTW:

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Consider the sequence $\{a_k\}_{k\ge 1}$ is defined by $a_1=1$, $a_2=\dfrac{1}{2}$ and $a_{k+2}=a_k+\dfrac{a_{k+1}}{2}+\dfrac{1}{4a_ka_{k+1}}$ for $k\ge 1.$

Prove that $\dfrac{1}{a_1a_3}+\dfrac{1}{a_2a_4}+\dfrac{1}{a_3a_5}+\cdots+\dfrac{1}{a_{98}a_{100}}<4$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[MarkFL]

Hello, MHB Community! (Wave)

anemone has asked me to fill in for her this week. :)

No one answered this week's problem, and the solution provided to me by anemone is as follows:

Spoiler

Note that $\dfrac{1}{a_ka_{k+2}}<\dfrac{2}{a_ka_{k+1}}-\dfrac{2}{a_{k+1}a_{k+2}}$, because it is equivalent to the inequality $a_{k+2}>a_k+\dfrac{1}{2}a_{k+1}$, which is a true fact derived from the given sequence.

Now we have
$\begin{align*}\dfrac{1}{a_1a_3}+\dfrac{1}{a_2a_4}+\dfrac{1}{a_3a_5}+\cdots+\dfrac{1}{a_{98}a_{100}}&<\dfrac{2}{a_1a_2}-\dfrac{2}{a_2a_3}+\dfrac{2}{a_2a_3}-\dfrac{2}{a_3a_4}+\cdots+\dfrac{2}{a_{97}a_{98}}-\dfrac{2}{a_{98}a_{99}}+\dfrac{2}{a_{98}a_{99}}-\dfrac{2}{a_{99}a_{100}}\\&<\dfrac{2}{a_1a_2}\\&=4\end{align*}$