Show $frac{n+1}{2n^2+2} in O(frac{1}{n})$ without limits












0












$begingroup$


$frac{n+1}{2n^2+2} in O(frac{1}{n})$ without limits



$frac{n+1}{2n^2+2} leq frac{n+n}{2n^2 + 2} leq frac{2n}{2n^2 + 2}$



Not sure how to lower the denominator






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are we allowed to use calculus? Because you can show the point (the $n_0$) where, with a sufficiently large $c$ (multiplying factor), that $frac{n + 1}{2n^2 + 2} leq cfrac{1}{n}$, for all $n > n_0$.
    $endgroup$
    – Jared
    Jan 31 at 3:17












  • $begingroup$
    And if not, you can just choose a fairly large $n_0$...it might also help to factor out $n$: $frac{n}{n^2}frac{1 + frac{1}{n}}{2 + frac{2}{n^2}}$.
    $endgroup$
    – Jared
    Jan 31 at 3:19










  • $begingroup$
    Note that $2n^2lt2n^2+2$
    $endgroup$
    – robjohn♦
    Jan 31 at 3:31


















0












$begingroup$


$frac{n+1}{2n^2+2} in O(frac{1}{n})$ without limits



$frac{n+1}{2n^2+2} leq frac{n+n}{2n^2 + 2} leq frac{2n}{2n^2 + 2}$



Not sure how to lower the denominator






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are we allowed to use calculus? Because you can show the point (the $n_0$) where, with a sufficiently large $c$ (multiplying factor), that $frac{n + 1}{2n^2 + 2} leq cfrac{1}{n}$, for all $n > n_0$.
    $endgroup$
    – Jared
    Jan 31 at 3:17












  • $begingroup$
    And if not, you can just choose a fairly large $n_0$...it might also help to factor out $n$: $frac{n}{n^2}frac{1 + frac{1}{n}}{2 + frac{2}{n^2}}$.
    $endgroup$
    – Jared
    Jan 31 at 3:19










  • $begingroup$
    Note that $2n^2lt2n^2+2$
    $endgroup$
    – robjohn♦
    Jan 31 at 3:31
















0












0








0





$begingroup$


$frac{n+1}{2n^2+2} in O(frac{1}{n})$ without limits



$frac{n+1}{2n^2+2} leq frac{n+n}{2n^2 + 2} leq frac{2n}{2n^2 + 2}$



Not sure how to lower the denominator






calculus







share|cite|improve this question











$endgroup$




$frac{n+1}{2n^2+2} in O(frac{1}{n})$ without limits



$frac{n+1}{2n^2+2} leq frac{n+n}{2n^2 + 2} leq frac{2n}{2n^2 + 2}$



Not sure how to lower the denominator






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 31 at 3:15







Bas bas

















asked Jan 31 at 3:10









Bas basBas bas

49512




49512












  • $begingroup$
    Are we allowed to use calculus? Because you can show the point (the $n_0$) where, with a sufficiently large $c$ (multiplying factor), that $frac{n + 1}{2n^2 + 2} leq cfrac{1}{n}$, for all $n > n_0$.
    $endgroup$
    – Jared
    Jan 31 at 3:17












  • $begingroup$
    And if not, you can just choose a fairly large $n_0$...it might also help to factor out $n$: $frac{n}{n^2}frac{1 + frac{1}{n}}{2 + frac{2}{n^2}}$.
    $endgroup$
    – Jared
    Jan 31 at 3:19










  • $begingroup$
    Note that $2n^2lt2n^2+2$
    $endgroup$
    – robjohn♦
    Jan 31 at 3:31




















  • $begingroup$
    Are we allowed to use calculus? Because you can show the point (the $n_0$) where, with a sufficiently large $c$ (multiplying factor), that $frac{n + 1}{2n^2 + 2} leq cfrac{1}{n}$, for all $n > n_0$.
    $endgroup$
    – Jared
    Jan 31 at 3:17












  • $begingroup$
    And if not, you can just choose a fairly large $n_0$...it might also help to factor out $n$: $frac{n}{n^2}frac{1 + frac{1}{n}}{2 + frac{2}{n^2}}$.
    $endgroup$
    – Jared
    Jan 31 at 3:19










  • $begingroup$
    Note that $2n^2lt2n^2+2$
    $endgroup$
    – robjohn♦
    Jan 31 at 3:31


















$begingroup$
Are we allowed to use calculus? Because you can show the point (the $n_0$) where, with a sufficiently large $c$ (multiplying factor), that $frac{n + 1}{2n^2 + 2} leq cfrac{1}{n}$, for all $n > n_0$.
$endgroup$
– Jared
Jan 31 at 3:17






$begingroup$
Are we allowed to use calculus? Because you can show the point (the $n_0$) where, with a sufficiently large $c$ (multiplying factor), that $frac{n + 1}{2n^2 + 2} leq cfrac{1}{n}$, for all $n > n_0$.
$endgroup$
– Jared
Jan 31 at 3:17














$begingroup$
And if not, you can just choose a fairly large $n_0$...it might also help to factor out $n$: $frac{n}{n^2}frac{1 + frac{1}{n}}{2 + frac{2}{n^2}}$.
$endgroup$
– Jared
Jan 31 at 3:19




$begingroup$
And if not, you can just choose a fairly large $n_0$...it might also help to factor out $n$: $frac{n}{n^2}frac{1 + frac{1}{n}}{2 + frac{2}{n^2}}$.
$endgroup$
– Jared
Jan 31 at 3:19












$begingroup$
Note that $2n^2lt2n^2+2$
$endgroup$
– robjohn♦
Jan 31 at 3:31






$begingroup$
Note that $2n^2lt2n^2+2$
$endgroup$
– robjohn♦
Jan 31 at 3:31












1 Answer
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$begingroup$

Note that $n + 1 le n + n = 2n$ for all $n ge 1$, so
$$
frac{n+1}{2n^2+2} le frac{n}{n^2+1}.
$$
Now notice that $n^2 + 1 > n^2$, so if $n ge 1$,
$$
frac{n+1}{2n^2+2} le frac{n}{n^2+1} < frac{n}{n^2} = frac{1}{n},
$$
as desired.






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    Note that $n + 1 le n + n = 2n$ for all $n ge 1$, so
    $$
    frac{n+1}{2n^2+2} le frac{n}{n^2+1}.
    $$
    Now notice that $n^2 + 1 > n^2$, so if $n ge 1$,
    $$
    frac{n+1}{2n^2+2} le frac{n}{n^2+1} < frac{n}{n^2} = frac{1}{n},
    $$
    as desired.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Note that $n + 1 le n + n = 2n$ for all $n ge 1$, so
      $$
      frac{n+1}{2n^2+2} le frac{n}{n^2+1}.
      $$
      Now notice that $n^2 + 1 > n^2$, so if $n ge 1$,
      $$
      frac{n+1}{2n^2+2} le frac{n}{n^2+1} < frac{n}{n^2} = frac{1}{n},
      $$
      as desired.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Note that $n + 1 le n + n = 2n$ for all $n ge 1$, so
        $$
        frac{n+1}{2n^2+2} le frac{n}{n^2+1}.
        $$
        Now notice that $n^2 + 1 > n^2$, so if $n ge 1$,
        $$
        frac{n+1}{2n^2+2} le frac{n}{n^2+1} < frac{n}{n^2} = frac{1}{n},
        $$
        as desired.






        share|cite|improve this answer









        $endgroup$



        Note that $n + 1 le n + n = 2n$ for all $n ge 1$, so
        $$
        frac{n+1}{2n^2+2} le frac{n}{n^2+1}.
        $$
        Now notice that $n^2 + 1 > n^2$, so if $n ge 1$,
        $$
        frac{n+1}{2n^2+2} le frac{n}{n^2+1} < frac{n}{n^2} = frac{1}{n},
        $$
        as desired.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 31 at 3:36









        Alex OrtizAlex Ortiz

        11.3k21441




        11.3k21441






























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