What is the maximum value of this fraction?












4












$begingroup$


If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






sequences-and-series algebra-precalculus means geometric-series







share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    47 mins ago


















4












$begingroup$


If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






sequences-and-series algebra-precalculus means geometric-series







share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    47 mins ago
















4












4








4


1



$begingroup$


If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






sequences-and-series algebra-precalculus means geometric-series







share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




If $x$ is positive, what is the maximum value of this expression:



$$frac{x^{100}}{1+x+x^2+ldots+x^{200}}$$



This question is from a book of problems on sequence and series under the section on AM-GM-HM inequality.



This is what I have tried:



The denominator is a geometric series whose sum is



$$frac{1-x^{201}}{1-x}$$



The fraction now becomes



$$frac{x^{100}(1-x)}{1-x^{201}}$$



I can imagine that solving this problem will require taking the AM/GM/HM of some expressions of $x$ and applying the AM-GM-HM inequality.



That means the above fractions should themselves be one of GM or HM (whose maximum value will be given by the corresponding AM and GM respectively).



I can't see such means from looking at the fraction. Can someone help me here?






sequences-and-series algebra-precalculus means geometric-series




sequences-and-series algebra-precalculus means geometric-series






share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









user69284user69284

876




876




New contributor




user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    47 mins ago
















  • 2




    $begingroup$
    For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
    $endgroup$
    – Mark Viola
    47 mins ago










2




2




$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
47 mins ago






$begingroup$
For $x>0$ we have$$sum_{n=0}^{200}x^nge 201 sqrt[201]{prod_{n=0}^{200}x^n}=201 sqrt[201]{x^{20100}}=201x^{100}$$ $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$
$endgroup$
– Mark Viola
47 mins ago












2 Answers
2






active

oldest

votes


















4












$begingroup$

The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    For $x>0$, we have from the AM-GM inequality



    $$begin{align}
    sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
    &=201 sqrt[201]{x^{20100}}\\
    &=201x^{100}
    end{align}$$



    Hence, we see that



    $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Please let me know how I can improve my answer. I really want to give you the best answer I can.
      $endgroup$
      – Mark Viola
      30 mins ago











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    user69284 is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3093099%2fwhat-is-the-maximum-value-of-this-fraction%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.






        share|cite|improve this answer









        $endgroup$



        The denominator (in the original form) is a multiple of an arithmetic mean - a sum of $201$ terms is $201$ times their average. So then, depending on taste, you can either apply AM-GM to the denominator or GM-HM to the whole thing.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 50 mins ago









        jmerryjmerry

        5,977718




        5,977718























            3












            $begingroup$

            For $x>0$, we have from the AM-GM inequality



            $$begin{align}
            sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
            &=201 sqrt[201]{x^{20100}}\\
            &=201x^{100}
            end{align}$$



            Hence, we see that



            $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              30 mins ago
















            3












            $begingroup$

            For $x>0$, we have from the AM-GM inequality



            $$begin{align}
            sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
            &=201 sqrt[201]{x^{20100}}\\
            &=201x^{100}
            end{align}$$



            Hence, we see that



            $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              30 mins ago














            3












            3








            3





            $begingroup$

            For $x>0$, we have from the AM-GM inequality



            $$begin{align}
            sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
            &=201 sqrt[201]{x^{20100}}\\
            &=201x^{100}
            end{align}$$



            Hence, we see that



            $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$






            share|cite|improve this answer









            $endgroup$



            For $x>0$, we have from the AM-GM inequality



            $$begin{align}
            sum_{n=0}^{200}x^n&ge 201 sqrt[201]{prod_{n=0}^{200}x^n}\\
            &=201 sqrt[201]{x^{20100}}\\
            &=201x^{100}
            end{align}$$



            Hence, we see that



            $$frac{x^{100}}{sum_{n=0}^{200}x^n}le frac1{201}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 44 mins ago









            Mark ViolaMark Viola

            131k1275171




            131k1275171












            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              30 mins ago


















            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can.
              $endgroup$
              – Mark Viola
              30 mins ago
















            $begingroup$
            Please let me know how I can improve my answer. I really want to give you the best answer I can.
            $endgroup$
            – Mark Viola
            30 mins ago




            $begingroup$
            Please let me know how I can improve my answer. I really want to give you the best answer I can.
            $endgroup$
            – Mark Viola
            30 mins ago










            user69284 is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            user69284 is a new contributor. Be nice, and check out our Code of Conduct.













            user69284 is a new contributor. Be nice, and check out our Code of Conduct.












            user69284 is a new contributor. Be nice, and check out our Code of Conduct.
















            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3093099%2fwhat-is-the-maximum-value-of-this-fraction%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            404 Error Contact Form 7 ajax form submitting

            Image
            0 The contact form does not work, and affects loading after one try. The chrome console shows the error message: POST http://example.com/wp-json/contact-form-7/v1/contact-forms/50/feedback 404 (Not Found) send @ jquery.js?ver=1.12.4:4 ajax @ jquery.js?ver=1.12.4:4 wpcf7.submit @ scripts.js?ver=5.0.3:346 (anonymous) @ scripts.js?ver=5.0.3:53 dispatch @ jquery.js?ver=1.12.4:3 r.handle @ jquery.js?ver=1.12.4:3 How to solve this issue? thanks in advance javascript wordpress contact-form-7 share | improve this question edited Jul 17 '18 at 5:39 Kiran Shahi 4,745 3 16 43
            Read more

            How to know if a Active Directory user can login interactively

            Image
            1 I would like to know if and how is it possible to know if a AD user can login interactively (on a server) in a domain. I need to know if I can find it out using a LDAP search. Thanks. windows active-directory login user-permissions user-profile share | improve this question edited 1 hour ago yagmoth555 ♦ 11.6k 3 17 42 asked 1 hour ago Luigi Luigi 6 1 New contributor
            Read more

            TypeError: fit_transform() missing 1 required positional argument: 'X'

            Image
            0 I am trying to do Feature Scaling in a dataset, but I get an error and have no idea how to proceed: > Traceback (most recent call last): > > File "<ipython-input-10-71bea414b4d0>", line 22, in <module> > x_train = sc_X.fit_transform(x_train) > > TypeError: fit_transform() missing 1 required positional argument: 'X' and here is my code: import pandas as pd # Importing the dataset dataset = pd.read_csv('Data.csv') X = dataset.iloc[:, :-1].values y = dataset.iloc[:, 3].values # Taking care of missing data from sklearn.preprocessing import Imputer imputer = Imputer(missing_values="NaN", strategy="mean", axis=0) imputer = Imputer.fit(imputer,X[:,1:3]) X[:, 1:3] = Imputer.transform(imputer,X[
            Read more