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Question:

Assmue that the postive integer $a,c$ ,such $\lfloor \sqrt{a} \rfloor=c$ ,Now let sequence $$y_{1}=1,y_{2}=-2c, y_{n+2}=-2c\cdot y_{n+1}+(a-c^2)y_{n},n\ge 1$$

show that $$T_{n}=y_{1}+y_{2}+\cdots+y_{n}\neq 0,\forall n\in N^{+}$$

where $\lfloor a \rfloor$ is the biggest integer not greater than $a$

My idea: since $$T_{n}-T_{n-1}=y_{n}$$ so $$T_{n+2}-T_{n+1}=-2c\cdot(T_{n+1}-T_{n})+(a-c^2)(T_{n}-T_{n-1})$$ so $$T_{n+2}=(1-2c)T_{n+1}+(2c+a-c^2)T_{n}-(a-c^2)T_{n-1}$$ then I can't prove $T_{n}\neq 0$

Thank you

math110
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1 Answers1

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First, obseve that the mapping $(y_n,T_n)\mapsto (-1)^{n-1} (y_n,T_n)$ brings the recurrence to

$$y_1=1;\; y_2=2c;\; y_{n+2}=2c y_{n+1}+(a-c^2)y_n\text{ for }n\geq 1$$

and the sum to the alternating series $$T_n=y_n-y_{n-1}+\cdots + (-1)^{n-1}y_1=y_n-T_{n-1}.$$ Note that $T_1=y_1=1>0$ trivially and $T_2=y_2-y_1=2c-1>0$ since $c$ is a positive integer.

Assume now that $T_{k}>0$ for some $k\geq 1$. Then \begin{align} T_{k+2} =y_{k+2}-y_{k+1}+T_k &> y_{k+2}-y_{k+1} &\text{(induction hypothesis)}\\ &=(2c-1)y_{k+1}+(a-c^2)y_k\\ &>(a-c^2)y_k &\text{($c$ is a positive integer)}\\ &\geq 0 &\text{($\sqrt{a}\geq c=\lfloor \sqrt{a}\rfloor$)}. \end{align} Therefore induction on the two base cases implies $T_n>0$ for all $n$. Since $T_n = 0$ iff $(-1)^{n-1} T_n = 0$, we conclude that none of the original sums can vanish either.

Semiclassical
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