$lim_{n to +infty} n cdotcos^2left(frac{n pi}{3}right)$












0












$begingroup$


Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$



First I have a look at $cos(frac{n pi}{3})$



What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?



$cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.



For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!






real-analysis limits limits-without-lhopital







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    0












    $begingroup$


    Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$



    First I have a look at $cos(frac{n pi}{3})$



    What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?



    $cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.



    For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!






    real-analysis limits limits-without-lhopital







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$



      First I have a look at $cos(frac{n pi}{3})$



      What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?



      $cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.



      For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!






      real-analysis limits limits-without-lhopital







      share|cite|improve this question











      $endgroup$




      Find $lim_{n to +infty} n cdot cos^2left(frac{n pi}{3}right)$



      First I have a look at $cos(frac{n pi}{3})$



      What I expect is that the sequence diverges so I want to find two sub-sequences that tend to a different value, correct?



      $cos$ is periodic for $2 pi$ so for the first sub-sequence let n:=6m and I get $cos(frac{6m cdot pi}{3})=cos(2m cdot pi)$ which is 1 for every m.



      For the next sub-sequence it seems to me that finding a k that gets me either $cos(3pi)$ or $cos(pi)$ would be a good idea since as then the sub-sequence will always deliver 0. However I get think of any k that satisfies that. Can someone help me out here? Thanks in advance!






      real-analysis limits limits-without-lhopital




      real-analysis limits limits-without-lhopital






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 22:31









      gimusi

      92.9k94494




      92.9k94494










      asked Dec 3 '18 at 22:20









      D. JohnD. John

      283




      283






















          2 Answers
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          $begingroup$

          Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
            $endgroup$
            – D. John
            Dec 3 '18 at 22:36










          • $begingroup$
            I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 8:27



















          0












          $begingroup$

          We have that $forall nin mathbb{N}$



          $$cos^2left(frac{n pi}{3}right) ge frac14$$



          therefore



          $$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

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            0












            $begingroup$

            Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
              $endgroup$
              – D. John
              Dec 3 '18 at 22:36










            • $begingroup$
              I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
              $endgroup$
              – Mostafa Ayaz
              Dec 4 '18 at 8:27
















            0












            $begingroup$

            Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
              $endgroup$
              – D. John
              Dec 3 '18 at 22:36










            • $begingroup$
              I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
              $endgroup$
              – Mostafa Ayaz
              Dec 4 '18 at 8:27














            0












            0








            0





            $begingroup$

            Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?






            share|cite|improve this answer









            $endgroup$



            Unfortunately, you cannot find two subsequences convergent to different values since the general limit is $infty$. Let $$a_{3n}=3ncos^2{npi}=3n\a_{3n+1}={(3n+1)}cos^2{(npi +{pi over 3})}={3n+1over 4}\a_{3n-1}={(3n-1)}cos^2{(npi -{pi over 3})}={3n+1over 4}$$what can you conclude from here?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 3 '18 at 22:26









            Mostafa AyazMostafa Ayaz

            15.4k3939




            15.4k3939












            • $begingroup$
              Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
              $endgroup$
              – D. John
              Dec 3 '18 at 22:36










            • $begingroup$
              I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
              $endgroup$
              – Mostafa Ayaz
              Dec 4 '18 at 8:27


















            • $begingroup$
              Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
              $endgroup$
              – D. John
              Dec 3 '18 at 22:36










            • $begingroup$
              I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
              $endgroup$
              – Mostafa Ayaz
              Dec 4 '18 at 8:27
















            $begingroup$
            Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
            $endgroup$
            – D. John
            Dec 3 '18 at 22:36




            $begingroup$
            Hm I'm not sure actually. Do you want me to apply the squeeze theorem?
            $endgroup$
            – D. John
            Dec 3 '18 at 22:36












            $begingroup$
            I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 8:27




            $begingroup$
            I just argued that since any term of the sequence (or even infinitely many terms of any arbitrary subsequence such as $a_{n^2}$) fall(s) in at least one $a_{3n+k}$ for $k=0,1,-1$ and all those subsequences tend to $infty$ so does the sequence itself.
            $endgroup$
            – Mostafa Ayaz
            Dec 4 '18 at 8:27











            0












            $begingroup$

            We have that $forall nin mathbb{N}$



            $$cos^2left(frac{n pi}{3}right) ge frac14$$



            therefore



            $$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              We have that $forall nin mathbb{N}$



              $$cos^2left(frac{n pi}{3}right) ge frac14$$



              therefore



              $$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                We have that $forall nin mathbb{N}$



                $$cos^2left(frac{n pi}{3}right) ge frac14$$



                therefore



                $$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$






                share|cite|improve this answer











                $endgroup$



                We have that $forall nin mathbb{N}$



                $$cos^2left(frac{n pi}{3}right) ge frac14$$



                therefore



                $$ n cdot cos^2left(frac{n pi}{3}right) ge frac n 4to infty$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 3 '18 at 22:28

























                answered Dec 3 '18 at 22:23









                gimusigimusi

                92.9k94494




                92.9k94494






























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