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So I have the sequence

$$\bigg(\dfrac{4n-1}{5n}\bigg) \qquad n \in \{1,2, \ldots \}$$

I want to show that this sequence is increasing and bounded above, so that I can use the bounded monotone convergence theorem. It is very obvious to me that this function is increasing and bounded, but I am not sure how to prove this. Usually when I see these types of problems they are phrased in a form such as $a_{n+1}=f(a_n)$ which I am comfortable work with. Is there any way to phrase my problem such that the $n+1$ term relates to the $n$ term?

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    $\begingroup$ Here, as is often the case, it helps to simplify your expression first. $\endgroup$ – Marc van Leeuwen Sep 29 '17 at 6:54
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You have $a_n = \frac{4}{5}-\frac{1}{5n} < \frac{4}{5}$

Because $\frac{1}{5n} > \frac{1}{5(n+1)}$, then $a_n < a_{n+1}$.

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Do you HAVE to do it using Weierstrass Monotone Convergence theorem? It's simpler to compute the limit: $$\lim\left( \frac{4n-1}{5n} \right)=\lim\left(\frac{4}{5}-\frac{1}{5n} \right)=\frac{4}{5}.$$

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If you insist on expressing $a_{n+1}$ as a function of $a_n$, you could say $$ a_{n+1} = \frac{15 a_n - 16}{25 a_n - 25}$$

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  • $\begingroup$ How does this help show either bounded or increasing? $\endgroup$ – marty cohen Sep 29 '17 at 2:21
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$a_n =\dfrac{4n-1}{5n} $.

Proceeding very naively,

$\begin{array}\\ a_{n+1}-a_n &=\dfrac{4(n+1)-1}{5(n+1)}-\dfrac{4n-1}{5n}\\ &=\dfrac{4n+3}{5n+5}-\dfrac{4n-1}{5n}\\ &=\dfrac{(4n+3)(5n)-(4n-1)(5n+5)}{(5n+5)5n}\\ &=\dfrac{(20n^2+15n)-(20n^2+15n-5)}{(5n+5)5n}\\ &=\dfrac{5}{(5n+5)5n}\\ &=\dfrac{1}{(n+1)5n}\\ &> 0\\ \end{array} $

so the terms are increasing.

Also, using the canonical telescoping series,

$a_{n+1}-a_n =\dfrac15\dfrac{1}{(n+1)n} =\dfrac15\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) $ and it telescopes.

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