Prove $a_n = sqrt{n+1}-sqrt{n}$ is monotone $n ge 0$.












1












$begingroup$



Prove $a_n = sqrt{n+1}-sqrt{n}$ is monotone $n ge 0$.




To be monotone it must be either increasing or decreasing, so:



$a_n ge a_{n-1}$ or $a_n le a_{n-1}$



$sqrt{n+1}-sqrt{n} ge? sqrt{n}-sqrt{n-1} $



$sqrt{n+1}+sqrt{n-1} ge? sqrt{n} + sqrt{n}$



I know that $sqrt{n+1} ge sqrt{n}$ and $sqrt{n} ge sqrt{n-1}$



But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.



Can you help me to figure out the solution?






analysis monotone-functions







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$endgroup$












  • $begingroup$
    If you calculate $a_0$ and $a_1$ (and possibly $a_2$ for good measure), you can easily see which way the sequence is not monotone.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:29


















1












$begingroup$



Prove $a_n = sqrt{n+1}-sqrt{n}$ is monotone $n ge 0$.




To be monotone it must be either increasing or decreasing, so:



$a_n ge a_{n-1}$ or $a_n le a_{n-1}$



$sqrt{n+1}-sqrt{n} ge? sqrt{n}-sqrt{n-1} $



$sqrt{n+1}+sqrt{n-1} ge? sqrt{n} + sqrt{n}$



I know that $sqrt{n+1} ge sqrt{n}$ and $sqrt{n} ge sqrt{n-1}$



But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.



Can you help me to figure out the solution?






analysis monotone-functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you calculate $a_0$ and $a_1$ (and possibly $a_2$ for good measure), you can easily see which way the sequence is not monotone.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:29
















1












1








1





$begingroup$



Prove $a_n = sqrt{n+1}-sqrt{n}$ is monotone $n ge 0$.




To be monotone it must be either increasing or decreasing, so:



$a_n ge a_{n-1}$ or $a_n le a_{n-1}$



$sqrt{n+1}-sqrt{n} ge? sqrt{n}-sqrt{n-1} $



$sqrt{n+1}+sqrt{n-1} ge? sqrt{n} + sqrt{n}$



I know that $sqrt{n+1} ge sqrt{n}$ and $sqrt{n} ge sqrt{n-1}$



But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.



Can you help me to figure out the solution?






analysis monotone-functions







share|cite|improve this question









$endgroup$





Prove $a_n = sqrt{n+1}-sqrt{n}$ is monotone $n ge 0$.




To be monotone it must be either increasing or decreasing, so:



$a_n ge a_{n-1}$ or $a_n le a_{n-1}$



$sqrt{n+1}-sqrt{n} ge? sqrt{n}-sqrt{n-1} $



$sqrt{n+1}+sqrt{n-1} ge? sqrt{n} + sqrt{n}$



I know that $sqrt{n+1} ge sqrt{n}$ and $sqrt{n} ge sqrt{n-1}$



But I can't make a statement about their sum since they have opposite signs that is one is greater or equal and the other is less or equal.



Can you help me to figure out the solution?






analysis monotone-functions




analysis monotone-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 20:26









Sargis IskandaryanSargis Iskandaryan

560112




560112












  • $begingroup$
    If you calculate $a_0$ and $a_1$ (and possibly $a_2$ for good measure), you can easily see which way the sequence is not monotone.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:29




















  • $begingroup$
    If you calculate $a_0$ and $a_1$ (and possibly $a_2$ for good measure), you can easily see which way the sequence is not monotone.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:29


















$begingroup$
If you calculate $a_0$ and $a_1$ (and possibly $a_2$ for good measure), you can easily see which way the sequence is not monotone.
$endgroup$
– Arthur
Dec 1 '18 at 20:29






$begingroup$
If you calculate $a_0$ and $a_1$ (and possibly $a_2$ for good measure), you can easily see which way the sequence is not monotone.
$endgroup$
– Arthur
Dec 1 '18 at 20:29












4 Answers
4






active

oldest

votes


















1












$begingroup$

Note that, for $ngeq 1$,
$$ (sqrt{n+1}+sqrt{n-1})^2 = n+1+2sqrt{n^2-1}+n-1 = 2n+2sqrt{n^2-1}leq 2n +2n=4n, $$
so that
$$ sqrt{n+1}+sqrt{n-1} leq 2sqrt{n} quad iffquad sqrt{n+1}- sqrt n leq sqrt n - sqrt{n-1}. $$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    hint



    use the conjugate



    $$a_n=frac{1}{sqrt{n+1}+sqrt{n}}=frac{1}{b_n}$$



    now prove tha $(b_n)$ is increasing and conclude that $(a_n)$ is decreasing.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
      $endgroup$
      – Sargis Iskandaryan
      Dec 1 '18 at 20:35










    • $begingroup$
      @SargisIskandaryan if the denominator increases, the fraction decreases.
      $endgroup$
      – hamam_Abdallah
      Dec 1 '18 at 20:37



















    1












    $begingroup$

    $$
    \a_{n+1}le a_{n}<=>
    \sqrt{n+1}+sqrt{n-1}le2sqrt{n}<=>
    \(sqrt{n+1}+sqrt{n-1})^2le(2sqrt{n})^2<=>
    \sqrt{(n-1)(n+1)}=sqrt{n^2-1}le sqrt{n^2}=n
    $$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Square to $2n+2sqrt{n^2-1}le 4n$.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        1












        $begingroup$

        Note that, for $ngeq 1$,
        $$ (sqrt{n+1}+sqrt{n-1})^2 = n+1+2sqrt{n^2-1}+n-1 = 2n+2sqrt{n^2-1}leq 2n +2n=4n, $$
        so that
        $$ sqrt{n+1}+sqrt{n-1} leq 2sqrt{n} quad iffquad sqrt{n+1}- sqrt n leq sqrt n - sqrt{n-1}. $$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Note that, for $ngeq 1$,
          $$ (sqrt{n+1}+sqrt{n-1})^2 = n+1+2sqrt{n^2-1}+n-1 = 2n+2sqrt{n^2-1}leq 2n +2n=4n, $$
          so that
          $$ sqrt{n+1}+sqrt{n-1} leq 2sqrt{n} quad iffquad sqrt{n+1}- sqrt n leq sqrt n - sqrt{n-1}. $$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Note that, for $ngeq 1$,
            $$ (sqrt{n+1}+sqrt{n-1})^2 = n+1+2sqrt{n^2-1}+n-1 = 2n+2sqrt{n^2-1}leq 2n +2n=4n, $$
            so that
            $$ sqrt{n+1}+sqrt{n-1} leq 2sqrt{n} quad iffquad sqrt{n+1}- sqrt n leq sqrt n - sqrt{n-1}. $$






            share|cite|improve this answer









            $endgroup$



            Note that, for $ngeq 1$,
            $$ (sqrt{n+1}+sqrt{n-1})^2 = n+1+2sqrt{n^2-1}+n-1 = 2n+2sqrt{n^2-1}leq 2n +2n=4n, $$
            so that
            $$ sqrt{n+1}+sqrt{n-1} leq 2sqrt{n} quad iffquad sqrt{n+1}- sqrt n leq sqrt n - sqrt{n-1}. $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 '18 at 20:35









            MisterRiemannMisterRiemann

            5,8451624




            5,8451624























                3












                $begingroup$

                hint



                use the conjugate



                $$a_n=frac{1}{sqrt{n+1}+sqrt{n}}=frac{1}{b_n}$$



                now prove tha $(b_n)$ is increasing and conclude that $(a_n)$ is decreasing.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
                  $endgroup$
                  – Sargis Iskandaryan
                  Dec 1 '18 at 20:35










                • $begingroup$
                  @SargisIskandaryan if the denominator increases, the fraction decreases.
                  $endgroup$
                  – hamam_Abdallah
                  Dec 1 '18 at 20:37
















                3












                $begingroup$

                hint



                use the conjugate



                $$a_n=frac{1}{sqrt{n+1}+sqrt{n}}=frac{1}{b_n}$$



                now prove tha $(b_n)$ is increasing and conclude that $(a_n)$ is decreasing.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
                  $endgroup$
                  – Sargis Iskandaryan
                  Dec 1 '18 at 20:35










                • $begingroup$
                  @SargisIskandaryan if the denominator increases, the fraction decreases.
                  $endgroup$
                  – hamam_Abdallah
                  Dec 1 '18 at 20:37














                3












                3








                3





                $begingroup$

                hint



                use the conjugate



                $$a_n=frac{1}{sqrt{n+1}+sqrt{n}}=frac{1}{b_n}$$



                now prove tha $(b_n)$ is increasing and conclude that $(a_n)$ is decreasing.






                share|cite|improve this answer











                $endgroup$



                hint



                use the conjugate



                $$a_n=frac{1}{sqrt{n+1}+sqrt{n}}=frac{1}{b_n}$$



                now prove tha $(b_n)$ is increasing and conclude that $(a_n)$ is decreasing.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 1 '18 at 20:40

























                answered Dec 1 '18 at 20:29









                hamam_Abdallahhamam_Abdallah

                38k21634




                38k21634












                • $begingroup$
                  $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
                  $endgroup$
                  – Sargis Iskandaryan
                  Dec 1 '18 at 20:35










                • $begingroup$
                  @SargisIskandaryan if the denominator increases, the fraction decreases.
                  $endgroup$
                  – hamam_Abdallah
                  Dec 1 '18 at 20:37


















                • $begingroup$
                  $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
                  $endgroup$
                  – Sargis Iskandaryan
                  Dec 1 '18 at 20:35










                • $begingroup$
                  @SargisIskandaryan if the denominator increases, the fraction decreases.
                  $endgroup$
                  – hamam_Abdallah
                  Dec 1 '18 at 20:37
















                $begingroup$
                $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
                $endgroup$
                – Sargis Iskandaryan
                Dec 1 '18 at 20:35




                $begingroup$
                $sqrt{n+2} + sqrt{n+1} ge? sqrt{n+1} + sqrt{n}$ => $sqrt{n+2} ge? sqrt{n}$ => increasing since $n+2 ge n$ but can you clarify how is that helping me?
                $endgroup$
                – Sargis Iskandaryan
                Dec 1 '18 at 20:35












                $begingroup$
                @SargisIskandaryan if the denominator increases, the fraction decreases.
                $endgroup$
                – hamam_Abdallah
                Dec 1 '18 at 20:37




                $begingroup$
                @SargisIskandaryan if the denominator increases, the fraction decreases.
                $endgroup$
                – hamam_Abdallah
                Dec 1 '18 at 20:37











                1












                $begingroup$

                $$
                \a_{n+1}le a_{n}<=>
                \sqrt{n+1}+sqrt{n-1}le2sqrt{n}<=>
                \(sqrt{n+1}+sqrt{n-1})^2le(2sqrt{n})^2<=>
                \sqrt{(n-1)(n+1)}=sqrt{n^2-1}le sqrt{n^2}=n
                $$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  $$
                  \a_{n+1}le a_{n}<=>
                  \sqrt{n+1}+sqrt{n-1}le2sqrt{n}<=>
                  \(sqrt{n+1}+sqrt{n-1})^2le(2sqrt{n})^2<=>
                  \sqrt{(n-1)(n+1)}=sqrt{n^2-1}le sqrt{n^2}=n
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    $$
                    \a_{n+1}le a_{n}<=>
                    \sqrt{n+1}+sqrt{n-1}le2sqrt{n}<=>
                    \(sqrt{n+1}+sqrt{n-1})^2le(2sqrt{n})^2<=>
                    \sqrt{(n-1)(n+1)}=sqrt{n^2-1}le sqrt{n^2}=n
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    $$
                    \a_{n+1}le a_{n}<=>
                    \sqrt{n+1}+sqrt{n-1}le2sqrt{n}<=>
                    \(sqrt{n+1}+sqrt{n-1})^2le(2sqrt{n})^2<=>
                    \sqrt{(n-1)(n+1)}=sqrt{n^2-1}le sqrt{n^2}=n
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 1 '18 at 20:41









                    Samvel SafaryanSamvel Safaryan

                    497111




                    497111























                        0












                        $begingroup$

                        Square to $2n+2sqrt{n^2-1}le 4n$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Square to $2n+2sqrt{n^2-1}le 4n$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Square to $2n+2sqrt{n^2-1}le 4n$.






                            share|cite|improve this answer









                            $endgroup$



                            Square to $2n+2sqrt{n^2-1}le 4n$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 1 '18 at 20:34









                            J.G.J.G.

                            24.1k22539




                            24.1k22539






























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