Conditions for which the sum of the product of sequences diverges












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Let $(C_k)_{k=1}^{infty} geq 0$ be non-decreasing sequence of natural numbers. I am trying to figure out when does the sum $$ sum_{k=1}^{infty} C_k 2^{-k} $$ diverge. From my understanding $C_k$ should be something of the form $2^{k}$ as we need to kill the decay of geometric series. Initially I thought $C_k$ ~ $2^k$ but then I can have $frac{2^k} {k}$ which diverges and is not comparable with $2^{k}$.



Any suggestions on lower bounds of $C_k$ would be greatly appreciated!






real-analysis analysis







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    $begingroup$


    Let $(C_k)_{k=1}^{infty} geq 0$ be non-decreasing sequence of natural numbers. I am trying to figure out when does the sum $$ sum_{k=1}^{infty} C_k 2^{-k} $$ diverge. From my understanding $C_k$ should be something of the form $2^{k}$ as we need to kill the decay of geometric series. Initially I thought $C_k$ ~ $2^k$ but then I can have $frac{2^k} {k}$ which diverges and is not comparable with $2^{k}$.



    Any suggestions on lower bounds of $C_k$ would be greatly appreciated!






    real-analysis analysis







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $(C_k)_{k=1}^{infty} geq 0$ be non-decreasing sequence of natural numbers. I am trying to figure out when does the sum $$ sum_{k=1}^{infty} C_k 2^{-k} $$ diverge. From my understanding $C_k$ should be something of the form $2^{k}$ as we need to kill the decay of geometric series. Initially I thought $C_k$ ~ $2^k$ but then I can have $frac{2^k} {k}$ which diverges and is not comparable with $2^{k}$.



      Any suggestions on lower bounds of $C_k$ would be greatly appreciated!






      real-analysis analysis







      share|cite|improve this question











      $endgroup$




      Let $(C_k)_{k=1}^{infty} geq 0$ be non-decreasing sequence of natural numbers. I am trying to figure out when does the sum $$ sum_{k=1}^{infty} C_k 2^{-k} $$ diverge. From my understanding $C_k$ should be something of the form $2^{k}$ as we need to kill the decay of geometric series. Initially I thought $C_k$ ~ $2^k$ but then I can have $frac{2^k} {k}$ which diverges and is not comparable with $2^{k}$.



      Any suggestions on lower bounds of $C_k$ would be greatly appreciated!






      real-analysis analysis




      real-analysis analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 20:30







      Shorty

















      asked Dec 13 '18 at 20:17









      ShortyShorty

      1438




      1438






















          2 Answers
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          $begingroup$

          You could apply the ratio or the root test and get the sufficient conditions for absolute convergence $limsup |C_{k+1}/C_k| <2$ and $limsupsqrt[k]{|C_k|}<2$, respectively.






          share|cite|improve this answer









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          • $begingroup$
            Isn't the ratio supposed to be 1?
            $endgroup$
            – mm-crj
            Dec 13 '18 at 20:39










          • $begingroup$
            Well, here we have $a_k = C_k2^{-k}$.
            $endgroup$
            – Dirk
            Dec 13 '18 at 20:40










          • $begingroup$
            aah yes, I see..
            $endgroup$
            – mm-crj
            Dec 13 '18 at 20:42



















          0












          $begingroup$

          I think the ratio test would do the trick:



          $$lim_{kto infty}left|frac{a_{k+1}}{a_k}right|=L$$



          The series diverges for $L>1$. For a more general case where the limit doesn't exist $liminfleft|frac{a_{k+1}}{a_k}right|>1$ is a sufficient condition for divergence.






          share|cite|improve this answer











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            You could apply the ratio or the root test and get the sufficient conditions for absolute convergence $limsup |C_{k+1}/C_k| <2$ and $limsupsqrt[k]{|C_k|}<2$, respectively.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Isn't the ratio supposed to be 1?
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:39










            • $begingroup$
              Well, here we have $a_k = C_k2^{-k}$.
              $endgroup$
              – Dirk
              Dec 13 '18 at 20:40










            • $begingroup$
              aah yes, I see..
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:42
















            2












            $begingroup$

            You could apply the ratio or the root test and get the sufficient conditions for absolute convergence $limsup |C_{k+1}/C_k| <2$ and $limsupsqrt[k]{|C_k|}<2$, respectively.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Isn't the ratio supposed to be 1?
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:39










            • $begingroup$
              Well, here we have $a_k = C_k2^{-k}$.
              $endgroup$
              – Dirk
              Dec 13 '18 at 20:40










            • $begingroup$
              aah yes, I see..
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:42














            2












            2








            2





            $begingroup$

            You could apply the ratio or the root test and get the sufficient conditions for absolute convergence $limsup |C_{k+1}/C_k| <2$ and $limsupsqrt[k]{|C_k|}<2$, respectively.






            share|cite|improve this answer









            $endgroup$



            You could apply the ratio or the root test and get the sufficient conditions for absolute convergence $limsup |C_{k+1}/C_k| <2$ and $limsupsqrt[k]{|C_k|}<2$, respectively.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 20:38









            DirkDirk

            8,8002447




            8,8002447












            • $begingroup$
              Isn't the ratio supposed to be 1?
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:39










            • $begingroup$
              Well, here we have $a_k = C_k2^{-k}$.
              $endgroup$
              – Dirk
              Dec 13 '18 at 20:40










            • $begingroup$
              aah yes, I see..
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:42


















            • $begingroup$
              Isn't the ratio supposed to be 1?
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:39










            • $begingroup$
              Well, here we have $a_k = C_k2^{-k}$.
              $endgroup$
              – Dirk
              Dec 13 '18 at 20:40










            • $begingroup$
              aah yes, I see..
              $endgroup$
              – mm-crj
              Dec 13 '18 at 20:42
















            $begingroup$
            Isn't the ratio supposed to be 1?
            $endgroup$
            – mm-crj
            Dec 13 '18 at 20:39




            $begingroup$
            Isn't the ratio supposed to be 1?
            $endgroup$
            – mm-crj
            Dec 13 '18 at 20:39












            $begingroup$
            Well, here we have $a_k = C_k2^{-k}$.
            $endgroup$
            – Dirk
            Dec 13 '18 at 20:40




            $begingroup$
            Well, here we have $a_k = C_k2^{-k}$.
            $endgroup$
            – Dirk
            Dec 13 '18 at 20:40












            $begingroup$
            aah yes, I see..
            $endgroup$
            – mm-crj
            Dec 13 '18 at 20:42




            $begingroup$
            aah yes, I see..
            $endgroup$
            – mm-crj
            Dec 13 '18 at 20:42











            0












            $begingroup$

            I think the ratio test would do the trick:



            $$lim_{kto infty}left|frac{a_{k+1}}{a_k}right|=L$$



            The series diverges for $L>1$. For a more general case where the limit doesn't exist $liminfleft|frac{a_{k+1}}{a_k}right|>1$ is a sufficient condition for divergence.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              I think the ratio test would do the trick:



              $$lim_{kto infty}left|frac{a_{k+1}}{a_k}right|=L$$



              The series diverges for $L>1$. For a more general case where the limit doesn't exist $liminfleft|frac{a_{k+1}}{a_k}right|>1$ is a sufficient condition for divergence.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                I think the ratio test would do the trick:



                $$lim_{kto infty}left|frac{a_{k+1}}{a_k}right|=L$$



                The series diverges for $L>1$. For a more general case where the limit doesn't exist $liminfleft|frac{a_{k+1}}{a_k}right|>1$ is a sufficient condition for divergence.






                share|cite|improve this answer











                $endgroup$



                I think the ratio test would do the trick:



                $$lim_{kto infty}left|frac{a_{k+1}}{a_k}right|=L$$



                The series diverges for $L>1$. For a more general case where the limit doesn't exist $liminfleft|frac{a_{k+1}}{a_k}right|>1$ is a sufficient condition for divergence.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 13 '18 at 20:47

























                answered Dec 13 '18 at 20:38









                mm-crjmm-crj

                425213




                425213






























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