Ink on Sequences #4

From the original condition, we see that

$(a_{n+2}-a_{n+1})n^2+2(a_{n+2}-a_{n+1})n+(-a_{n+1}+a_n)n+(-a_{n+1}+a_n)=0 \\ (n^2+2n)(a_{n+2}-a_{n+1})=(n+1)(a_{n+1}-a_n)$

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Note that $a_2\neq a_1$ .

If $a_{n+1}\neq a_n$ ,

$(n^2+2n)(a_{n+2}-a_{n+1})=(n+1)(a_{n+1}-a_n)\neq0$ .

$\therefore~a_{n+2}\neq a_{n+1}$ .

Therefore, using above induction, we conclude that for every positive integer $n$ , $a_{n+1}\neq a_n$ .

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Then we can transform the equation.

$\begin{aligned} \frac{a_{n+2}-a_{n+1}}{a_{n+1}-a_n} & =\frac{n+1}{n(n+2)} \\ \prod_{k=1}^{n-1} \frac{a_{k+2}-a_{k+1}}{a_{k+1}-a_k} & =\prod_{k=1}^{n-1} \frac{k+1}{k(k+2)} \end{aligned}$

Note that $n$ must be larger than or equal to $2$ .

$\begin{aligned} \frac{a_{n+1}-a_n}{a_2-a_1}&=\frac{n!}{(n-1)!\cdot\dfrac{(n+1)!}{2}} \quad (n\geq2)\\ a_{n+1}-a_n&=144\cdot\frac{n}{(n+1)!} \quad (n\geq2) \end{aligned}$

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Now let's get some sum.

$\begin{aligned} \sum_{k=2}^{n-1} (a_{k+1}-a_k)&=144\sum_{k=2}^{n-1} \frac{k}{(k+1)!} \quad (n\geq3) \\ a_n-a_2&=144\sum_{k=2}^{n-1} \left(\frac{1}{k!}-\frac{1}{(k+1)!}\right) \quad (n\geq3) \\ a_n&=121+144\left(\frac{1}{2}-\frac{1}{n!}\right) \quad (n\geq3) \end{aligned}$

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So we've got $a_n$ .

Then let's punch that into the limit we're trying to get a value of.

$\lim_{n\rightarrow\infty} a_n=\lim_{n\rightarrow\infty} \left(121+144\left(\frac{1}{2}-\frac{1}{n!}\right)\right)=121+72=\boxed{193}$

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The answer is already found, but I'd like to elaborate more.

Substitute $n=1,~2$ to $a_n=121+144\left(\frac{1}{2}-\frac{1}{n!}\right)$ .

And we see that the equation holds.

Therefore, for every positive integer $n$ ,

$a_n=121+144\left(\frac{1}{2}-\frac{1}{n!}\right)$