Determining the convergence of $ sumlimits_{n=1}^{infty}frac{-(n-1)}{nsqrt{n+1}}$












2












$begingroup$


We have to determin the convergence of the following sum:
begin{equation}
sumlimits_{n=1}^{infty}frac{-(n-1)}{nsqrt{n+1}}
end{equation}



And we've seen multiple ways to determine if the sum converges or diverges, but they only work with non negative terms. So tests like d'Alemberts ratio test, Cauchy's root test or Raabe's test wont work. How would I go about determining the convergence of this sum?






calculus convergence summation







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$endgroup$












  • $begingroup$
    then get rid of the $-1$, since $sum frac{1}{n^{3/2}}$ converges. so you have $sum_n frac{1}{sqrt{n}}$, which diverges
    $endgroup$
    – mathworker21
    Jan 7 at 16:54










  • $begingroup$
    @CalvinGodfrey I don't think that's "without loss of generality"
    $endgroup$
    – mathworker21
    Jan 7 at 16:55
















2












$begingroup$


We have to determin the convergence of the following sum:
begin{equation}
sumlimits_{n=1}^{infty}frac{-(n-1)}{nsqrt{n+1}}
end{equation}



And we've seen multiple ways to determine if the sum converges or diverges, but they only work with non negative terms. So tests like d'Alemberts ratio test, Cauchy's root test or Raabe's test wont work. How would I go about determining the convergence of this sum?






calculus convergence summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    then get rid of the $-1$, since $sum frac{1}{n^{3/2}}$ converges. so you have $sum_n frac{1}{sqrt{n}}$, which diverges
    $endgroup$
    – mathworker21
    Jan 7 at 16:54










  • $begingroup$
    @CalvinGodfrey I don't think that's "without loss of generality"
    $endgroup$
    – mathworker21
    Jan 7 at 16:55














2












2








2





$begingroup$


We have to determin the convergence of the following sum:
begin{equation}
sumlimits_{n=1}^{infty}frac{-(n-1)}{nsqrt{n+1}}
end{equation}



And we've seen multiple ways to determine if the sum converges or diverges, but they only work with non negative terms. So tests like d'Alemberts ratio test, Cauchy's root test or Raabe's test wont work. How would I go about determining the convergence of this sum?






calculus convergence summation







share|cite|improve this question











$endgroup$




We have to determin the convergence of the following sum:
begin{equation}
sumlimits_{n=1}^{infty}frac{-(n-1)}{nsqrt{n+1}}
end{equation}



And we've seen multiple ways to determine if the sum converges or diverges, but they only work with non negative terms. So tests like d'Alemberts ratio test, Cauchy's root test or Raabe's test wont work. How would I go about determining the convergence of this sum?






calculus convergence summation




calculus convergence summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 2:23









Key Flex

7,78461232




7,78461232










asked Jan 7 at 16:51









ViktorViktor

1389




1389












  • $begingroup$
    then get rid of the $-1$, since $sum frac{1}{n^{3/2}}$ converges. so you have $sum_n frac{1}{sqrt{n}}$, which diverges
    $endgroup$
    – mathworker21
    Jan 7 at 16:54










  • $begingroup$
    @CalvinGodfrey I don't think that's "without loss of generality"
    $endgroup$
    – mathworker21
    Jan 7 at 16:55


















  • $begingroup$
    then get rid of the $-1$, since $sum frac{1}{n^{3/2}}$ converges. so you have $sum_n frac{1}{sqrt{n}}$, which diverges
    $endgroup$
    – mathworker21
    Jan 7 at 16:54










  • $begingroup$
    @CalvinGodfrey I don't think that's "without loss of generality"
    $endgroup$
    – mathworker21
    Jan 7 at 16:55
















$begingroup$
then get rid of the $-1$, since $sum frac{1}{n^{3/2}}$ converges. so you have $sum_n frac{1}{sqrt{n}}$, which diverges
$endgroup$
– mathworker21
Jan 7 at 16:54




$begingroup$
then get rid of the $-1$, since $sum frac{1}{n^{3/2}}$ converges. so you have $sum_n frac{1}{sqrt{n}}$, which diverges
$endgroup$
– mathworker21
Jan 7 at 16:54












$begingroup$
@CalvinGodfrey I don't think that's "without loss of generality"
$endgroup$
– mathworker21
Jan 7 at 16:55




$begingroup$
@CalvinGodfrey I don't think that's "without loss of generality"
$endgroup$
– mathworker21
Jan 7 at 16:55










3 Answers
3






active

oldest

votes


















2












$begingroup$

You can use Integral test:



$$f(n)=dfrac{n-1}{nsqrt{n+1}}$$



When you plot a graph, you will notice that $f(n)$ is positive, continuous and decreasing from $n=4$



So, $$int_4^infty dfrac{n-1}{nsqrt{n+1}}=mbox{diverges}$$



So, by integral test, the series diverges






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
    $endgroup$
    – Viktor
    Jan 7 at 16:59








  • 1




    $begingroup$
    @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
    $endgroup$
    – Key Flex
    Jan 7 at 17:03






  • 2




    $begingroup$
    Okay thanks i think i understand now!
    $endgroup$
    – Viktor
    Jan 7 at 17:03



















3












$begingroup$

For $n>2$,
$$frac{n-1}{nsqrt{n+1}}>frac1{2sqrt{n+1}} $$
As $sum frac1{2sqrt{n+1}}$ diverges. so does $sumfrac{n-1}{nsqrt{n+1}}$ and also $sumfrac{-(n-1)}{nsqrt{n+1}}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You may also use the comparison test.



    Intuitively, as $n$ gets larger, $n-1$ is essentially $n$, and $sqrt{n+1}$ is essentially $sqrt{n}$. That means you can compare $frac{n-1}{nsqrt{n+1}}$ with $frac{n}{nsqrt{n}}=frac{1}{sqrt{n}}$



    Formally, you can find a constant $C>0$ such that:
    $$frac{n-1}{nsqrt{n+1}}>Cfrac{1}{sqrt{n}}, text{ for }ngeq 2$$



    (what could that constant be?)



    Now since $sum_{n=1}^inftyfrac{1}{sqrt{n}}$ diverges, that means $sum_{n=1}^inftyfrac{n-1}{nsqrt{n+1}}$ diverges.



    Obs.: For any non-zero constant $lambda$, we have that $sum_{n=1}^infty lambda, a_n$ converges $Leftrightarrow sum_{n=1}^infty a_n$ converges (this should be a simple exercise). In particular, the negative sign in $frac{-(n-1)}{nsqrt{n+1}}$ is not relevant to convergence).






    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









      2












      $begingroup$

      You can use Integral test:



      $$f(n)=dfrac{n-1}{nsqrt{n+1}}$$



      When you plot a graph, you will notice that $f(n)$ is positive, continuous and decreasing from $n=4$



      So, $$int_4^infty dfrac{n-1}{nsqrt{n+1}}=mbox{diverges}$$



      So, by integral test, the series diverges






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
        $endgroup$
        – Viktor
        Jan 7 at 16:59








      • 1




        $begingroup$
        @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
        $endgroup$
        – Key Flex
        Jan 7 at 17:03






      • 2




        $begingroup$
        Okay thanks i think i understand now!
        $endgroup$
        – Viktor
        Jan 7 at 17:03
















      2












      $begingroup$

      You can use Integral test:



      $$f(n)=dfrac{n-1}{nsqrt{n+1}}$$



      When you plot a graph, you will notice that $f(n)$ is positive, continuous and decreasing from $n=4$



      So, $$int_4^infty dfrac{n-1}{nsqrt{n+1}}=mbox{diverges}$$



      So, by integral test, the series diverges






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
        $endgroup$
        – Viktor
        Jan 7 at 16:59








      • 1




        $begingroup$
        @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
        $endgroup$
        – Key Flex
        Jan 7 at 17:03






      • 2




        $begingroup$
        Okay thanks i think i understand now!
        $endgroup$
        – Viktor
        Jan 7 at 17:03














      2












      2








      2





      $begingroup$

      You can use Integral test:



      $$f(n)=dfrac{n-1}{nsqrt{n+1}}$$



      When you plot a graph, you will notice that $f(n)$ is positive, continuous and decreasing from $n=4$



      So, $$int_4^infty dfrac{n-1}{nsqrt{n+1}}=mbox{diverges}$$



      So, by integral test, the series diverges






      share|cite|improve this answer









      $endgroup$



      You can use Integral test:



      $$f(n)=dfrac{n-1}{nsqrt{n+1}}$$



      When you plot a graph, you will notice that $f(n)$ is positive, continuous and decreasing from $n=4$



      So, $$int_4^infty dfrac{n-1}{nsqrt{n+1}}=mbox{diverges}$$



      So, by integral test, the series diverges







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 7 at 16:56









      Key FlexKey Flex

      7,78461232




      7,78461232












      • $begingroup$
        I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
        $endgroup$
        – Viktor
        Jan 7 at 16:59








      • 1




        $begingroup$
        @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
        $endgroup$
        – Key Flex
        Jan 7 at 17:03






      • 2




        $begingroup$
        Okay thanks i think i understand now!
        $endgroup$
        – Viktor
        Jan 7 at 17:03


















      • $begingroup$
        I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
        $endgroup$
        – Viktor
        Jan 7 at 16:59








      • 1




        $begingroup$
        @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
        $endgroup$
        – Key Flex
        Jan 7 at 17:03






      • 2




        $begingroup$
        Okay thanks i think i understand now!
        $endgroup$
        – Viktor
        Jan 7 at 17:03
















      $begingroup$
      I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
      $endgroup$
      – Viktor
      Jan 7 at 16:59






      $begingroup$
      I have 2 questions about this: first, why did you take out the negative sign? Can you take out the negative sign and when you do this and the remaining part converges or diverges, does it still converge or diverges, and second why can you start at 4 and not at 1 in your integral?
      $endgroup$
      – Viktor
      Jan 7 at 16:59






      1




      1




      $begingroup$
      @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
      $endgroup$
      – Key Flex
      Jan 7 at 17:03




      $begingroup$
      @Viktor first, I took the constant out, here $(-1)$, and then performed the test. This doesn't change the result. I took the boundary from $4$ only after checking if $f(n)$ is positive, continuous and decreasing.
      $endgroup$
      – Key Flex
      Jan 7 at 17:03




      2




      2




      $begingroup$
      Okay thanks i think i understand now!
      $endgroup$
      – Viktor
      Jan 7 at 17:03




      $begingroup$
      Okay thanks i think i understand now!
      $endgroup$
      – Viktor
      Jan 7 at 17:03











      3












      $begingroup$

      For $n>2$,
      $$frac{n-1}{nsqrt{n+1}}>frac1{2sqrt{n+1}} $$
      As $sum frac1{2sqrt{n+1}}$ diverges. so does $sumfrac{n-1}{nsqrt{n+1}}$ and also $sumfrac{-(n-1)}{nsqrt{n+1}}$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        For $n>2$,
        $$frac{n-1}{nsqrt{n+1}}>frac1{2sqrt{n+1}} $$
        As $sum frac1{2sqrt{n+1}}$ diverges. so does $sumfrac{n-1}{nsqrt{n+1}}$ and also $sumfrac{-(n-1)}{nsqrt{n+1}}$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          For $n>2$,
          $$frac{n-1}{nsqrt{n+1}}>frac1{2sqrt{n+1}} $$
          As $sum frac1{2sqrt{n+1}}$ diverges. so does $sumfrac{n-1}{nsqrt{n+1}}$ and also $sumfrac{-(n-1)}{nsqrt{n+1}}$.






          share|cite|improve this answer









          $endgroup$



          For $n>2$,
          $$frac{n-1}{nsqrt{n+1}}>frac1{2sqrt{n+1}} $$
          As $sum frac1{2sqrt{n+1}}$ diverges. so does $sumfrac{n-1}{nsqrt{n+1}}$ and also $sumfrac{-(n-1)}{nsqrt{n+1}}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 16:58









          Hagen von EitzenHagen von Eitzen

          277k22269496




          277k22269496























              1












              $begingroup$

              You may also use the comparison test.



              Intuitively, as $n$ gets larger, $n-1$ is essentially $n$, and $sqrt{n+1}$ is essentially $sqrt{n}$. That means you can compare $frac{n-1}{nsqrt{n+1}}$ with $frac{n}{nsqrt{n}}=frac{1}{sqrt{n}}$



              Formally, you can find a constant $C>0$ such that:
              $$frac{n-1}{nsqrt{n+1}}>Cfrac{1}{sqrt{n}}, text{ for }ngeq 2$$



              (what could that constant be?)



              Now since $sum_{n=1}^inftyfrac{1}{sqrt{n}}$ diverges, that means $sum_{n=1}^inftyfrac{n-1}{nsqrt{n+1}}$ diverges.



              Obs.: For any non-zero constant $lambda$, we have that $sum_{n=1}^infty lambda, a_n$ converges $Leftrightarrow sum_{n=1}^infty a_n$ converges (this should be a simple exercise). In particular, the negative sign in $frac{-(n-1)}{nsqrt{n+1}}$ is not relevant to convergence).






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                You may also use the comparison test.



                Intuitively, as $n$ gets larger, $n-1$ is essentially $n$, and $sqrt{n+1}$ is essentially $sqrt{n}$. That means you can compare $frac{n-1}{nsqrt{n+1}}$ with $frac{n}{nsqrt{n}}=frac{1}{sqrt{n}}$



                Formally, you can find a constant $C>0$ such that:
                $$frac{n-1}{nsqrt{n+1}}>Cfrac{1}{sqrt{n}}, text{ for }ngeq 2$$



                (what could that constant be?)



                Now since $sum_{n=1}^inftyfrac{1}{sqrt{n}}$ diverges, that means $sum_{n=1}^inftyfrac{n-1}{nsqrt{n+1}}$ diverges.



                Obs.: For any non-zero constant $lambda$, we have that $sum_{n=1}^infty lambda, a_n$ converges $Leftrightarrow sum_{n=1}^infty a_n$ converges (this should be a simple exercise). In particular, the negative sign in $frac{-(n-1)}{nsqrt{n+1}}$ is not relevant to convergence).






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You may also use the comparison test.



                  Intuitively, as $n$ gets larger, $n-1$ is essentially $n$, and $sqrt{n+1}$ is essentially $sqrt{n}$. That means you can compare $frac{n-1}{nsqrt{n+1}}$ with $frac{n}{nsqrt{n}}=frac{1}{sqrt{n}}$



                  Formally, you can find a constant $C>0$ such that:
                  $$frac{n-1}{nsqrt{n+1}}>Cfrac{1}{sqrt{n}}, text{ for }ngeq 2$$



                  (what could that constant be?)



                  Now since $sum_{n=1}^inftyfrac{1}{sqrt{n}}$ diverges, that means $sum_{n=1}^inftyfrac{n-1}{nsqrt{n+1}}$ diverges.



                  Obs.: For any non-zero constant $lambda$, we have that $sum_{n=1}^infty lambda, a_n$ converges $Leftrightarrow sum_{n=1}^infty a_n$ converges (this should be a simple exercise). In particular, the negative sign in $frac{-(n-1)}{nsqrt{n+1}}$ is not relevant to convergence).






                  share|cite|improve this answer











                  $endgroup$



                  You may also use the comparison test.



                  Intuitively, as $n$ gets larger, $n-1$ is essentially $n$, and $sqrt{n+1}$ is essentially $sqrt{n}$. That means you can compare $frac{n-1}{nsqrt{n+1}}$ with $frac{n}{nsqrt{n}}=frac{1}{sqrt{n}}$



                  Formally, you can find a constant $C>0$ such that:
                  $$frac{n-1}{nsqrt{n+1}}>Cfrac{1}{sqrt{n}}, text{ for }ngeq 2$$



                  (what could that constant be?)



                  Now since $sum_{n=1}^inftyfrac{1}{sqrt{n}}$ diverges, that means $sum_{n=1}^inftyfrac{n-1}{nsqrt{n+1}}$ diverges.



                  Obs.: For any non-zero constant $lambda$, we have that $sum_{n=1}^infty lambda, a_n$ converges $Leftrightarrow sum_{n=1}^infty a_n$ converges (this should be a simple exercise). In particular, the negative sign in $frac{-(n-1)}{nsqrt{n+1}}$ is not relevant to convergence).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 15 at 14:56

























                  answered Jan 7 at 17:12









                  rmdmc89rmdmc89

                  2,1071922




                  2,1071922






























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