Series - $sum_{i=1}^infty (frac{5}{12})^i$ - geometric series?












2












$begingroup$


I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57


















2












$begingroup$


I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57
















2












2








2





$begingroup$


I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$




I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion




calculus sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 14:04









Larry

2,41331129




2,41331129










asked Jan 16 at 13:57









AlanAlan

1,3841021




1,3841021








  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57
















  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57










4




4




$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57






$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57












3 Answers
3






active

oldest

votes


















5












$begingroup$

HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
    $endgroup$
    – Alan
    Jan 16 at 14:01










  • $begingroup$
    exactly correct @Alan
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:01










  • $begingroup$
    Thank you for your answer and for your time.
    $endgroup$
    – Alan
    Jan 16 at 14:02










  • $begingroup$
    no problem @Alan .. you can mark the answer as correct if you found it useful :)
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:03






  • 1




    $begingroup$
    Yes, in a few minutes when I can :-)
    $endgroup$
    – Alan
    Jan 16 at 14:03



















1












$begingroup$

There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, a great point of view.
    $endgroup$
    – Alan
    Jan 16 at 14:23



















1












$begingroup$

Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






share|cite|improve this answer









$endgroup$













    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075757%2fseries-sum-i-1-infty-frac512i-geometric-series%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03
















    5












    $begingroup$

    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03














    5












    5








    5





    $begingroup$

    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






    share|cite|improve this answer









    $endgroup$



    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 13:59









    Ahmad BazziAhmad Bazzi

    8,3282824




    8,3282824








    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03














    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03








    3




    3




    $begingroup$
    which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
    $endgroup$
    – Alan
    Jan 16 at 14:01




    $begingroup$
    which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
    $endgroup$
    – Alan
    Jan 16 at 14:01












    $begingroup$
    exactly correct @Alan
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:01




    $begingroup$
    exactly correct @Alan
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:01












    $begingroup$
    Thank you for your answer and for your time.
    $endgroup$
    – Alan
    Jan 16 at 14:02




    $begingroup$
    Thank you for your answer and for your time.
    $endgroup$
    – Alan
    Jan 16 at 14:02












    $begingroup$
    no problem @Alan .. you can mark the answer as correct if you found it useful :)
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:03




    $begingroup$
    no problem @Alan .. you can mark the answer as correct if you found it useful :)
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:03




    1




    1




    $begingroup$
    Yes, in a few minutes when I can :-)
    $endgroup$
    – Alan
    Jan 16 at 14:03




    $begingroup$
    Yes, in a few minutes when I can :-)
    $endgroup$
    – Alan
    Jan 16 at 14:03











    1












    $begingroup$

    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23
















    1












    $begingroup$

    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23














    1












    1








    1





    $begingroup$

    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






    share|cite|improve this answer









    $endgroup$



    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 14:20









    BernardBernard

    121k740116




    121k740116












    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23


















    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23
















    $begingroup$
    Thank you, a great point of view.
    $endgroup$
    – Alan
    Jan 16 at 14:23




    $begingroup$
    Thank you, a great point of view.
    $endgroup$
    – Alan
    Jan 16 at 14:23











    1












    $begingroup$

    Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






        share|cite|improve this answer









        $endgroup$



        Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 14:40









        drhabdrhab

        102k545136




        102k545136






























            draft saved

            draft discarded




















































            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%2f3075757%2fseries-sum-i-1-infty-frac512i-geometric-series%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

            'app-layout' is not a known element: how to share Component with different Modules

            Image
            0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
            Read more

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
            Read more

            WPF add header to Image with URL pettitions [duplicate]

            Image
            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
            Read more