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Let $Y = \{y_n\}$ be defined inductively by $y_1=1$ , $y_{n+1} = \frac 14\left(2y_n +3\right)$. Show that $$\lim_{n\to \infty}y_n=\frac 32$$

This is a problem from Bartle's Introduction to Real analysis. Specifically 3.3.4 Example A

The book walks us through the proof, and I understand the inductive steps needed to prove that $y_n < 2$ and that $y_n< y_{n+1}$.

But then, when calculating the limit, the book examines the 1-Tail of the series and notes that there is an algebraic relation between the nth term of the series and the nth term of the 1-Tail series. This is where I am stuck. What is this relationship and how is it used to calculate the limit.

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  • $\begingroup$ What series are you referring to? $\endgroup$ – Umberto P. Apr 19 '17 at 20:48
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$$ y_{n+1} = \frac 14\left(2y_n +3\right) $$ so $$ y_{n+1}-\frac32 = \frac 14\left(2y_n +3\right) - \frac32 \\ = \frac12(y_n-\frac32) $$

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