Ink on Sequences #5

Let $a_n=dn+k.$

Then $a_{2n}=2dn+k,$ which, for some real number $p,$ leads to $a_2+a_4+a_6+~\cdots~+a_{2n}=dn^2+pn.$

We can also see that $a_{2n-1}=2dn-2d+k=2dn+l,$ which, for some real number $q,$ leads to $a_1+a_3+a_5+~\cdots~+a_{2n-1}=dn^2+qn.$

Subtract $a_1+a_3+a_5+~\cdots~+a_{2n-1}=dn^2+qn$ from $a_2+a_4+a_6+~\cdots~+a_{2n}=dn^2+pn$ and the result is:

$dn=(p-q)n.$ So $p-d=q,$ which leads to $a_1+a_3+a_5+~\cdots~+a_{2n-1}=dn^2+(p-d)n.$

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$\begin{aligned} &\lim_{n\to\infty} \left(\sqrt{a_2+a_4+a_6+~\cdots~+a_{2n}}-\sqrt{a_1+a_3+a_5+~\cdots~+a_{2n-1}}\right) \\ &=\lim_{n\to\infty} \left(\sqrt{dn^2+pn}-\sqrt{dn^2+(p-d)n}\right) \\ &=\lim_{n\to\infty} \frac{\left(\sqrt{dn^2+pn+q}-\sqrt{dn^2+(p-d)n+q}\right)\left(\sqrt{dn^2+pn+q}+\sqrt{dn^2+(p-d)n+q}\right)}{\sqrt{dn^2+pn+q}+\sqrt{dn^2+(p-d)n+q}} \\ &=\lim_{n\to\infty} \frac{dn^2+pn-dn^2-(p-d)n}{2\sqrt{d}n} \\ &=\lim_{n\to\infty} \frac{dn}{2\sqrt{d}n} \\ &=\frac{\sqrt{d}}{2} \end{aligned}$

In this problem, $\sqrt{2017}-\sqrt{2016}=d.$

Therefore, the value we're trying to get is $\dfrac{\left(\dfrac{\sqrt{d}}{2}\right)^2}{d}=\boxed{\dfrac{1}{4}}.$