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Here's the question: Let $t_1 = 1$ and $t_{n+1} = \left[1-\frac{1}{(n+1)^2}\right] t_n \ \ \forall n \in \mathbb{N}$. Prove that lim $t_n$ exists.

Here's my proof. Can someone please verify it or offer suggestions for improvement?

$\textbf{Proof:}$ First, we prove two simple facts about $(t_n):$

(1) $\forall n \in \mathbb{N}$, $t_n \geqslant 0$

To see this, let $\phi_n$ be the statement that $t_n \geqslant 0$. Then, $\phi_1$ is true, since $t_1 = 1$.

Now, let $\phi_n$ be true for some $n \in \mathbb{N}$. Then, $t_n \geqslant 0$.

Then, $t_{n+1} = \left[1-\frac{1}{(n+1)^2}\right] t_n$. Since $t_n \geqslant 0$ by the inductive hypothesis, and $\left[1-\frac{1}{(n+1)^2}\right] \geqslant 0 \ \ \forall n \in \mathbb{N}$, $t_{n+1} \geqslant 0$.

Therefore, $t_n \geqslant 0 \ \ \forall n \in \mathbb{N}$

(2) $\forall n \in \mathbb{N}$, $t_{n+1} \leqslant t_n \leqslant 1$

To see this, let $\psi_n$ be the statement that $t_{n+1} \leqslant t_n \leqslant 1$. Then, $\psi_1$ is true, since $t_1 = 1$ and $t_2 = \frac{3}{4}$

Then, $t_{n+2} = \left[1-\frac{1}{(n+1)^2}\right]t_{n+1}$.

Now, $\left[1-\frac{1}{(n+1)^2}\right] \leqslant 1 \ \ \forall n \in \mathbb{N}$, and $t_{n+1} \leqslant t_n$, by the inductive hypothesis. Therefore, $t_{n+2} \leqslant t_{n+1} \leqslant 1$.

Therefore, $\forall n \in \mathbb{N}$, $t_{n+1} \leqslant t_n \leqslant 1$.

Hence, $\forall n \in \mathbb{N}$, $t_n \in [0, 1]$. Hence, $t_n$ is bounded. Also, $t_n$ is decreasing, by (2). Therefore, by the monotone convergence theorem, lim $t_n$ exists.

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    $\begingroup$ Looks fine. Perhaps you could be more specific and remark that the sequence is bounded below and monotone decreasing and thus its limits exists (and equals it infimum). $\endgroup$ – DonAntonio May 7 '14 at 19:23
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Your proof looks OK. Another way to prove that the limit exists would be to simply calculate it. Notice that

$$ t_n = \left (1 - \frac{1}{n^2} \right)t_{n-1} = \frac{(n+1)(n-1)}{n^2}t_{n-1} $$ Thus we have:

$$ t_n = \frac{(n+1)(n-1)}{n^2} \frac{n(n-2)}{(n-1)^2} \frac{(n-1)(n-3)}{(n-2)^2} \frac{(n-2)(n-4)}{(n-3)^2} \dots \frac{(4)(2)}{3^2} \frac{(3)(1)}{2^2} t_1 $$

Everything except $\frac{n+1}{n}$ from the left and $\frac12$ from the right reduces. Substituting $t_1=1$ leads to:

$$ t_n = \frac{n+1}{2n} $$

The limit of which is $\frac12$.

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