Determine if the infinite series converge or diverge












0














The series is as follows:



begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?



I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.



Any hint on how to get started on this one is appreciated.
Thanks in prior






calculus sequences-and-series







share|cite|improve this question


















  • 1




    Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
    – Alex
    Dec 3 '18 at 20:10










  • Thank you very much Alex. I didn't see this one coming.
    – Allorja
    Dec 3 '18 at 20:23
















0














The series is as follows:



begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?



I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.



Any hint on how to get started on this one is appreciated.
Thanks in prior






calculus sequences-and-series







share|cite|improve this question


















  • 1




    Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
    – Alex
    Dec 3 '18 at 20:10










  • Thank you very much Alex. I didn't see this one coming.
    – Allorja
    Dec 3 '18 at 20:23














0












0








0







The series is as follows:



begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?



I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.



Any hint on how to get started on this one is appreciated.
Thanks in prior






calculus sequences-and-series







share|cite|improve this question













The series is as follows:



begin{equation}
sum {frac{n-2}{nsqrt{4n+2}}}
end{equation}
Which test suits the best for determing the convergence/divergence of this series?



I tried the Ratio test,the Limit comparison with ${frac{1}{n}}$ and with ${frac{1}{n^2}}$, I can't seem to find a "bigger" series that converges to use the Direct Comparison test and all of them with no result. I can't use the root test.



Any hint on how to get started on this one is appreciated.
Thanks in prior






calculus sequences-and-series




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 Dec 3 '18 at 20:06









AllorjaAllorja

499




499








  • 1




    Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
    – Alex
    Dec 3 '18 at 20:10










  • Thank you very much Alex. I didn't see this one coming.
    – Allorja
    Dec 3 '18 at 20:23














  • 1




    Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
    – Alex
    Dec 3 '18 at 20:10










  • Thank you very much Alex. I didn't see this one coming.
    – Allorja
    Dec 3 '18 at 20:23








1




1




Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10




Compare the summand to $n^{-frac{1}{2}}$. The series diverges.
– Alex
Dec 3 '18 at 20:10












Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23




Thank you very much Alex. I didn't see this one coming.
– Allorja
Dec 3 '18 at 20:23










3 Answers
3






active

oldest

votes


















3














Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of



$$sum frac{n-2}{nsqrt {4n+2}},$$



think about



$$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$



Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.






share|cite|improve this answer





















  • This just made my calculus easier for the rest of the semester. Thank you sir!
    – Allorja
    Dec 4 '18 at 15:56



















1














For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.






share|cite|improve this answer





















  • +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
    – Allorja
    Dec 3 '18 at 20:25



















0














In my opinion the simpler way is to note that



$$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$



and then refer to limit comparison test with $sum frac1{sqrt n}$.






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of



    $$sum frac{n-2}{nsqrt {4n+2}},$$



    think about



    $$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$



    Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.






    share|cite|improve this answer





















    • This just made my calculus easier for the rest of the semester. Thank you sir!
      – Allorja
      Dec 4 '18 at 15:56
















    3














    Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of



    $$sum frac{n-2}{nsqrt {4n+2}},$$



    think about



    $$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$



    Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.






    share|cite|improve this answer





















    • This just made my calculus easier for the rest of the semester. Thank you sir!
      – Allorja
      Dec 4 '18 at 15:56














    3












    3








    3






    Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of



    $$sum frac{n-2}{nsqrt {4n+2}},$$



    think about



    $$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$



    Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.






    share|cite|improve this answer












    Here's a way to think about this: Constants don't mean anything compared to the growth of $n^p$ for any $p>0.$ Thus instead of



    $$sum frac{n-2}{nsqrt {4n+2}},$$



    think about



    $$sum frac{n}{nsqrt {n}} = sum frac{1}{sqrt {n}}.$$



    Since the last series diverges, your guess should be "divergence" for the original series. That's the way calculus experts think about this. You still have to make sure your guess is correct, but instead of wandering around in the darkness, you have a good guide for the "first guess," and that is often the hardest part.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 4 '18 at 0:40









    zhw.zhw.

    71.7k43075




    71.7k43075












    • This just made my calculus easier for the rest of the semester. Thank you sir!
      – Allorja
      Dec 4 '18 at 15:56


















    • This just made my calculus easier for the rest of the semester. Thank you sir!
      – Allorja
      Dec 4 '18 at 15:56
















    This just made my calculus easier for the rest of the semester. Thank you sir!
    – Allorja
    Dec 4 '18 at 15:56




    This just made my calculus easier for the rest of the semester. Thank you sir!
    – Allorja
    Dec 4 '18 at 15:56











    1














    For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.






    share|cite|improve this answer





















    • +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
      – Allorja
      Dec 3 '18 at 20:25
















    1














    For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.






    share|cite|improve this answer





















    • +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
      – Allorja
      Dec 3 '18 at 20:25














    1












    1








    1






    For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.






    share|cite|improve this answer












    For $nge 4$, $n-2gefrac{n}{2}$ while $4n+2lefrac{9n}{2}$, so $frac{n-2}{nsqrt{4n+2}}gefrac{1}{sqrt{18n}}$. This confirms divergence.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 20:10









    J.G.J.G.

    23.4k22137




    23.4k22137












    • +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
      – Allorja
      Dec 3 '18 at 20:25


















    • +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
      – Allorja
      Dec 3 '18 at 20:25
















    +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
    – Allorja
    Dec 3 '18 at 20:25




    +1 definitely and thank you for your help. But comparing with $n^-0.5$ seems clearer for me.
    – Allorja
    Dec 3 '18 at 20:25











    0














    In my opinion the simpler way is to note that



    $$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$



    and then refer to limit comparison test with $sum frac1{sqrt n}$.






    share|cite|improve this answer


























      0














      In my opinion the simpler way is to note that



      $$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$



      and then refer to limit comparison test with $sum frac1{sqrt n}$.






      share|cite|improve this answer
























        0












        0








        0






        In my opinion the simpler way is to note that



        $$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$



        and then refer to limit comparison test with $sum frac1{sqrt n}$.






        share|cite|improve this answer












        In my opinion the simpler way is to note that



        $$frac{n-2}{nsqrt{4n+2}}simfrac{n}{nsqrt{4n}}=frac1{2sqrt n}$$



        and then refer to limit comparison test with $sum frac1{sqrt n}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 20:25









        gimusigimusi

        1




        1






























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