Ink on Sequences #3

Let us first work out the recurrence relation of $\{a_n\}$ .

$\begin{aligned} a_n + S_n & = 2n & ...(1) \\ a_{n+1} + S_{n+1} & = 2(n+1) & ...(2) \\ (2)-(1): \quad a_{n+1} - a_n + S_{n+1} - S_n & = 2(n+1) - 2n \\ a_{n+1} - a_n + a_{n+1} & = 2 \\ \implies a_{n+1} & = \frac {a_n}2+1 & ...(3) \end{aligned}$

Form (1): $a_n + S_n = 2n \implies a_1 + S_1 = 2 \implies a_1 + a_1 = 2 \implies a_1 = 1$ . Then using (3):

$\begin{aligned} a_{\color{#D61F06}1} & = 1 & = \frac {2^{\color{#D61F06}1}-1}{2^{{\color{#D61F06}1}-1}} \\ a_{\color{#D61F06}2} & = \frac {a_1}2 +1 = \frac 32 & = \frac {2^{\color{#D61F06}2}-1}{2^{{\color{#D61F06}2}-1}} \\ a_{\color{#D61F06}3} & = \frac 34 +1 = \frac 74 & = \frac {2^{\color{#D61F06}3}-1}{2^{{\color{#D61F06}3}-1}} \end{aligned}$

From the first few $n$ , it appears that we can claim that $a_n = \dfrac {2^n-1}{2^{n-1}}$ . Let us prove by induction that the claim is true for all $n \ge 1$ .

Proof:

For $n=1$ , $a_1 = \dfrac {2^1-1}{2^{1-1}} = 1$ . Therefore, the claim is true for $n=1$ .

Assuming the claim is true for $n$ , then from (3):

$\begin{aligned} a_{\color{#D61F06}n+1} & = \frac {a_n}2+1 \\ & = \frac {2^n-1}{2^{n-1} \cdot2} + 1 \\ & = \frac {2^n - 1 + 2^n}{2^n} \\ & = \frac {2^{\color{#D61F06}n+1}-1}{2^{{\color{#D61F06}n+1}-1}} \end{aligned}$

Therefore, the claim is also true for $n+1$ and hence true for all $n \ge 1$ . $\square$

Now, we have:

$\begin{aligned} \lim_{n \to \infty} a_n & = \lim_{n \to \infty} \frac {2^n - 1}{2^{n-1}} \\ & = \lim_{n \to \infty} \left(2 - \frac 1{2^{n-1}}\right) \\ & = \boxed{2} \end{aligned}$

Firstly, note that: $\ (a_{n+1} +S_{n+1}) - (a_{n} +S_{n}) = 2$ reads $\ 2a_{n} - a_{n-1} = 2 \quad (1)$ Also, $\ a_{1} = 1$ now from (1) we can calculate successive terms to obtain $\ a_{2} = \frac{3}{2} , a_{3} = \frac{7}{4} \quad etc$ . To which I conjecture that the sequence is strictly monotonic increasing and bounded above by 2.

Claim 1: $\ a_{n} < 2$ Proof: By induction (using (1)) $\ a_{n+1} = 1+ \frac{a_{n}}{2} < 1 + \frac{2}{2} = 2$

Claim 2: $\ a_{n+1} > a_{n}$ Proof: By induction again: $\ a_{n+1} - a_{n} = 1 - \frac{a_{n}}{2} > 0$

Hence, by the Monotonic Sequences Theorem, $\ a_{n}$ converges, to say $\ a \in \mathbb{R}$ . That is $\ a_{n} \rightarrow a \ as \ n \rightarrow\infty$ . Therefore, by (1): $\ \lim_{n \rightarrow \infty } ( 2 a_{n} - a_{n-1} ) = 2a - a = a = 2$

(Note: I'm aware that I did skip a few details in the induction )

Let us assume $\{a_{n}\}$ is a convergent sequence . Hence we can use Cauchy's First Limit Theorem which states that the limit of the sequence $\{a_{n}\}$ is equal to the limit of the Arithmetic mean of the terms of the sequence provided the limit of the sequence exists.

Hence $\displaystyle \lim_{n\to\infty}\frac{S_{n}}{n} = \lim_{n\to\infty}a_{n}$ .

So using the second equation :-

$\displaystyle \lim_{n\to\infty}\left(\frac{a_{n}}{n} + \frac{S_{n}}{n}\right)= 2$ .

But since $a_{n}$ is assumed to be convergent it must be necessarily bounded. So $\displaystyle \lim_{n\to\infty}\frac{a_{n}}{n} = 0$ .

Hence $\displaystyle \lim_{n\to\infty}\frac{S_{n}}{n}=2$ .

But using the above theorem....this limit is the same as the limit of the sequence .

Hence $\displaystyle \lim_{n\to\infty}a_{n}=2$ .

Note:- I have assumed the convergence of $\{a_{n}\}$ since it was asked in the question. Otherwise we can also proceed as @Chew-Seong Cheong did to establish the recurrence relation $2a_{n+1}=a_{n} +2$ and then use this to prove the convergence of the sequence.