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$X=(x_n)$ defined as $X_1=1,X_2=2$ and $X_n=\dfrac12\left(X_{n-2}+X_{n-1}\right)$. To finds its limit, first we should prove that it is convergent, but my book has not given its solution. Instead, it just states that it is bounded between $1$ and $2$, i.e., $1< X_n < 2$ and guides to do it myself. It also states that $X_n$ is not monotone.

After some calculations, we find that mod of $X_n - X_{n-1}=\frac{1}{2^{n-1}}$. Please suggest how to proceed in showing that its Cauchy hence also convergent thanks in advance!

spaceisdarkgreen
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Vivek
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    Better than the modulus, after some calculation, we find that $X_n-X_{n-1} = (-2)^{-(n-2)}.$ And since we can sum differences up to get an exact value of $X_n,$ there's really no need to prove it's a Cauchy sequence separately. – spaceisdarkgreen Aug 18 '18 at 06:42

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$\sum _{n=1}^{\infty} |X_n-X_{n-1}| <\infty$ implies that $\{X_n\}$ is Cauchy. To see this take $k>m$ and note that $|X_k-X_{m-1}|=|\sum _{n=m}^{k} (X_n-X_{n-1})| \leq\sum _{n=m}^{k} |X_n-X_{n-1}| $. This last quantity tends to $0$ because teh series $\sum _{n=1}^{\infty} |X_n-X_{n-1}|$ is convergent. [ In fact $\sum _{n=m}^{k} |X_n-X_{n-1}|=S_k-S_{m-1}$ where $S_n$ is the n-th partial sum of the series; if $S_n \to s$ then $S_k-S_{m-1} \to s-s=0$ as $k>m \to \infty$]. We have proved that $\{X_n\}$ is Cauchy.

Kavi Rama Murthy
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