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If $S_n=\frac{1}{2}({S_{n-1}+S_{n-2}}), \forall n>2,$ then show that ${S_n}$ converges.

I have proved that sequence converges and I understood all the steps except the last one. First, I assumed that the $S_1<S_2$ which leads to a sequence such that it has two subsequences. One of even order and another of odd order. The odd ordered subsequence is increasing and bounded above by $S_2$ and hence it is convergent and similarly the even ordered subsequence also converges. Also, both subsequence converge to the same limit. We can check this by simple substitution of two different expression in the above sequence and then both expression will turn out to be same.

I understood all the steps of the book but then the following expression,I couldn't understand:

$S_{k}+ \frac {1}{2}S_{k-1}=\frac{1}{2}(S_1+2S_2).$

I couldn't not understand,how the book came to the above mentioned step. Any help or hint would be great. Thanks in advance

Guy Fsone
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userNoOne
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    The formula for $S_n$ can be readily derived since there's a linear recurrence relation, and the convergence of ${S_n}$ follows. – Ѕᴀᴀᴅ Jan 26 '18 at 04:07
  • Thanks Alex, I am not able to understand how the relationship in the last expression has been established. Would be great, if you can Help. – userNoOne Jan 26 '18 at 04:15
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    $$S_n = \frac{1}{2}(S_{n - 1} + S_{n - 2}) \Rightarrow S_n + \frac{1}{2}S_{n - 1} = S_{n - 1} + \frac{1}{2}S_{n - 2},$$so ${S_n + \frac{1}{2}S_{n - 1}}$ is a constant sequence. – Ѕᴀᴀᴅ Jan 26 '18 at 04:20
  • I understood that now. Thank you so much Alex. – userNoOne Jan 26 '18 at 04:21
  • $S_n = S_1+ \frac{2 }3(S_2-S_1)\left(1-\frac{(-1)^{n-1}}{2^{n-1}}\right) \to \color{red}{S_1+ \frac{2 }3(S_2-S_1)}$ – Guy Fsone Feb 02 '18 at 15:50
  • Another way is to directly use the recursive relation, and induction on $n\geq 1,$ to prove that $|S_n-S_{n-1}|=2^{1-n}|S_1-S_0|$ for all $n\geq 1.$ Then for $0\leq m <n$ we have $|S_m-S_n|=|\sum_{j=1}^{n-m}S_{m+j-1}-S_{m+i}|$ $\leq \sum_{j=1}^{n-m}|S_{m+j-1}-S_{m+j}|=$ $=\sum_{j=1}^{n-m}(2^{1-m-j})<2^{1-m},$ implying that the equence $(S_n)_n$ satisfies the Cauchy Criterion. – DanielWainfleet Feb 02 '18 at 19:18
  • The last 2 expressions in my comment above should have been multiplied by $|S_1-S_0|$. – DanielWainfleet Feb 02 '18 at 19:43

2 Answers2

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From the recurrence equation,

$$S_k+\frac12 S_{k-1}=\frac12(S_{k-1}+S_{k-2})+\frac12 S_{k-1}=S_{k-1}+\frac12S_{k-2},$$ and by induction,

$$S_k+\frac12 S_{k-1}=S_2+\frac12S_1.$$

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Answer: $$S_n = S_1+ \frac{2 }3(S_2-S_1)\left(1-\frac{(-1)^{n-1}}{2^{n-1}}\right) \to \color{red}{S_1+ \frac{2 }3(S_2-S_1)}$$

see below

Enforcing $X_n=S_n-S_{n-1}$ then, $$S_n=\frac{1}{2}({S_{n-1}+S_{n-2}}), \forall n>2\Longleftrightarrow S_n -S_{n-1}=-\frac{1}{2}({S_{n-1}-S_{n-2}}), \forall n>2,$$ that is $$X_n=-\frac{1}{2}X_{n-1}, \forall n>2,$$

which is geometric sequence with ratio $-\frac12$. Hence $$X_n =\frac{(-1)^{n-2}}{2^{n-2}}X_{2}~~~or ~~~S_n-S_{n-1}= \frac{(-1)^{n-2}}{2^{n-2}}X_{2}$$

Finally by telescopic sequence like here we get $$S_n -S_1= \sum_{k=2}^{n}(S_k -S_{k-1}) = X_{2}\sum_{k=2}^{n} \frac{(-1)^{k-2}}{2^{k-2}} =\frac{2X_{2}}3\left(1-\frac{(-1)^{n-1}}{2^{n-1}}\right)$$

Guy Fsone
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