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show by induction that the sequence : enter image description here

is monotonically increasing ... any help? it's a bit hard for me to write the induction hypothesis, I know that we must show that :

        $ a_1\le a_2$
        l
      Let it be for every natural n:
         $a_n\le a_{n+1}$ ----> hypothesis 
        and we must prove that it is also correct for $n+2$:
        $a_{n+1}\le a_{n+2}$ 

First what I said is correct? and I don't know how to continue from here... Any help is appreciated.

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We don't need to use induction, we just need to show $a_n < a_{n+1}$.

Let's start with \begin{eqnarray*} 2 &<& 3 \\ 4n+2 &<& 4n+3 \\ 2(2n+1) &<& 2n+2 + 2n+1 \end{eqnarray*} divide by $2(n+1)(2n+1)$ and we have \begin{eqnarray*} \frac{1}{n+1} < \frac{1}{2n+1} + \frac{1}{2(n+1)} \end{eqnarray*} add $ \frac{1}{n+2} +\cdots + \frac{1}{2n}$ to both sides, so \begin{eqnarray*} \frac{1}{n+1} +\cdots + \frac{1}{2n} < \frac{1}{n+2} + \cdots + \frac{1}{2(n+1)}. \end{eqnarray*} Which means $a_n <a_{n+1}$.

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  • $\begingroup$ but that is my hypnosis no ? and i must show that an+1<an+2 .. $\endgroup$ – Majd Dec 10 '19 at 21:18
  • $\begingroup$ Just run the argument backwards to show the hypothesis. $\endgroup$ – Donald Splutterwit Dec 10 '19 at 21:20
  • $\begingroup$ i still don't understand exactly what is the hypothesis and what i should show , can u edit it and type it again ? and show how u reached your final answer ? $\endgroup$ – Majd Dec 10 '19 at 21:23

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