Find upper bound for sequence [duplicate]












0












$begingroup$



This question already has an answer here:




  • How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$

    5 answers




$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$



I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.










share|cite|improve this question









$endgroup$



marked as duplicate by Martin R, Community Feb 3 at 10:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
    $endgroup$
    – Martin R
    Feb 3 at 10:43












  • $begingroup$
    Also related: math.stackexchange.com/q/525749/42969.
    $endgroup$
    – Martin R
    Feb 3 at 10:44
















0












$begingroup$



This question already has an answer here:




  • How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$

    5 answers




$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$



I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.










share|cite|improve this question









$endgroup$



marked as duplicate by Martin R, Community Feb 3 at 10:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
    $endgroup$
    – Martin R
    Feb 3 at 10:43












  • $begingroup$
    Also related: math.stackexchange.com/q/525749/42969.
    $endgroup$
    – Martin R
    Feb 3 at 10:44














0












0








0





$begingroup$



This question already has an answer here:




  • How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$

    5 answers




$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$



I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.










share|cite|improve this question









$endgroup$





This question already has an answer here:




  • How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$

    5 answers




$$ Fn = frac{1}{n+1} +frac{1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$$



I need to show that 1 is an upper bound for this sequence.I tried to use induction but didn't manage to prove it.





This question already has an answer here:




  • How do I prove $frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$

    5 answers








calculus sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 3 at 10:39









Nick KaklNick Kakl

176




176




marked as duplicate by Martin R, Community Feb 3 at 10:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Martin R, Community Feb 3 at 10:46


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
    $endgroup$
    – Martin R
    Feb 3 at 10:43












  • $begingroup$
    Also related: math.stackexchange.com/q/525749/42969.
    $endgroup$
    – Martin R
    Feb 3 at 10:44


















  • $begingroup$
    Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
    $endgroup$
    – Martin R
    Feb 3 at 10:43












  • $begingroup$
    Also related: math.stackexchange.com/q/525749/42969.
    $endgroup$
    – Martin R
    Feb 3 at 10:44
















$begingroup$
Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
$endgroup$
– Martin R
Feb 3 at 10:43






$begingroup$
Possible duplicate of How do I prove$frac 34geq frac{1}{n+1}+frac {1}{n+2}+frac{1}{n+3}+cdots+frac{1}{n+n}$
$endgroup$
– Martin R
Feb 3 at 10:43














$begingroup$
Also related: math.stackexchange.com/q/525749/42969.
$endgroup$
– Martin R
Feb 3 at 10:44




$begingroup$
Also related: math.stackexchange.com/q/525749/42969.
$endgroup$
– Martin R
Feb 3 at 10:44










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.






        share|cite|improve this answer









        $endgroup$



        Hint: observe that each term in the sum is bounded from above by $frac{1}{n}$ and there are $n$ terms.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 3 at 10:43









        HarnakHarnak

        1,3691512




        1,3691512















            Popular posts from this blog

            MongoDB - Not Authorized To Execute Command

            in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

            Npm cannot find a required file even through it is in the searched directory