Divergence of series $sum frac{k}{2k+1}$












1














Consider the infinite series denoted by:
$$sum frac{k}{2k+1}$$
starting at some arbitrary index. We wish to prove that this diverges. We will see what the limit of the general term $a_n$ is, if this does not go to zero, the series diverges by the limit test.
$$lim_{k rightarrow infty} frac{k}{2k+1}=lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}= frac{1}{2+0}=frac{1}{2}$$
By using the fraction limit theorem $(lim frac{A}{B}= frac{lim A}{lim B})$ and sum limit theorem respectively. We conclude that the series diverges as $a_n
not rightarrow 0$.



Is this correct?
I wanted to also try the following proof for $kgeq 1$:



$$ frac{k}{2k+1} > frac{1}{2k+1} geq frac{1}{2k+k} = frac{1}{3} frac{1}{k} $$
Now we simply get the harmonic series, but every factor is multiplied by $frac{1}{3}$, is there still some result we can use about multiplying divergent series by nonzero nonnegative constants?






calculus real-analysis sequences-and-series proof-verification







share|cite|improve this question




















  • 1




    That first proof looks good, and is how I would solve this problem.
    – Dave
    Nov 27 '18 at 22:22












  • Both ways are fine, but what do you mean with "the fraction limit theorem"?
    – gimusi
    Nov 27 '18 at 22:25










  • One of the standard results for limits is that if we take a limit of a fraction of two sequences, it is the same as the fraction of limits.
    – Wesley Strik
    Nov 27 '18 at 22:27






  • 1




    @WesleyGroupshaveFeelingsToo Ah ok you are referring to that one $$ lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}$$ That's fine!
    – gimusi
    Nov 27 '18 at 22:28












  • In my book a friendly introduction to analysis by Kosmala, it is referred to 2.2.1C
    – Wesley Strik
    Nov 27 '18 at 22:28
















1














Consider the infinite series denoted by:
$$sum frac{k}{2k+1}$$
starting at some arbitrary index. We wish to prove that this diverges. We will see what the limit of the general term $a_n$ is, if this does not go to zero, the series diverges by the limit test.
$$lim_{k rightarrow infty} frac{k}{2k+1}=lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}= frac{1}{2+0}=frac{1}{2}$$
By using the fraction limit theorem $(lim frac{A}{B}= frac{lim A}{lim B})$ and sum limit theorem respectively. We conclude that the series diverges as $a_n
not rightarrow 0$.



Is this correct?
I wanted to also try the following proof for $kgeq 1$:



$$ frac{k}{2k+1} > frac{1}{2k+1} geq frac{1}{2k+k} = frac{1}{3} frac{1}{k} $$
Now we simply get the harmonic series, but every factor is multiplied by $frac{1}{3}$, is there still some result we can use about multiplying divergent series by nonzero nonnegative constants?






calculus real-analysis sequences-and-series proof-verification







share|cite|improve this question




















  • 1




    That first proof looks good, and is how I would solve this problem.
    – Dave
    Nov 27 '18 at 22:22












  • Both ways are fine, but what do you mean with "the fraction limit theorem"?
    – gimusi
    Nov 27 '18 at 22:25










  • One of the standard results for limits is that if we take a limit of a fraction of two sequences, it is the same as the fraction of limits.
    – Wesley Strik
    Nov 27 '18 at 22:27






  • 1




    @WesleyGroupshaveFeelingsToo Ah ok you are referring to that one $$ lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}$$ That's fine!
    – gimusi
    Nov 27 '18 at 22:28












  • In my book a friendly introduction to analysis by Kosmala, it is referred to 2.2.1C
    – Wesley Strik
    Nov 27 '18 at 22:28














1












1








1







Consider the infinite series denoted by:
$$sum frac{k}{2k+1}$$
starting at some arbitrary index. We wish to prove that this diverges. We will see what the limit of the general term $a_n$ is, if this does not go to zero, the series diverges by the limit test.
$$lim_{k rightarrow infty} frac{k}{2k+1}=lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}= frac{1}{2+0}=frac{1}{2}$$
By using the fraction limit theorem $(lim frac{A}{B}= frac{lim A}{lim B})$ and sum limit theorem respectively. We conclude that the series diverges as $a_n
not rightarrow 0$.



Is this correct?
I wanted to also try the following proof for $kgeq 1$:



$$ frac{k}{2k+1} > frac{1}{2k+1} geq frac{1}{2k+k} = frac{1}{3} frac{1}{k} $$
Now we simply get the harmonic series, but every factor is multiplied by $frac{1}{3}$, is there still some result we can use about multiplying divergent series by nonzero nonnegative constants?






calculus real-analysis sequences-and-series proof-verification







share|cite|improve this question















Consider the infinite series denoted by:
$$sum frac{k}{2k+1}$$
starting at some arbitrary index. We wish to prove that this diverges. We will see what the limit of the general term $a_n$ is, if this does not go to zero, the series diverges by the limit test.
$$lim_{k rightarrow infty} frac{k}{2k+1}=lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}= frac{1}{2+0}=frac{1}{2}$$
By using the fraction limit theorem $(lim frac{A}{B}= frac{lim A}{lim B})$ and sum limit theorem respectively. We conclude that the series diverges as $a_n
not rightarrow 0$.



Is this correct?
I wanted to also try the following proof for $kgeq 1$:



$$ frac{k}{2k+1} > frac{1}{2k+1} geq frac{1}{2k+k} = frac{1}{3} frac{1}{k} $$
Now we simply get the harmonic series, but every factor is multiplied by $frac{1}{3}$, is there still some result we can use about multiplying divergent series by nonzero nonnegative constants?






calculus real-analysis sequences-and-series proof-verification




calculus real-analysis sequences-and-series proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 22:23







Wesley Strik

















asked Nov 27 '18 at 22:20









Wesley StrikWesley Strik

1,623423




1,623423








  • 1




    That first proof looks good, and is how I would solve this problem.
    – Dave
    Nov 27 '18 at 22:22












  • Both ways are fine, but what do you mean with "the fraction limit theorem"?
    – gimusi
    Nov 27 '18 at 22:25










  • One of the standard results for limits is that if we take a limit of a fraction of two sequences, it is the same as the fraction of limits.
    – Wesley Strik
    Nov 27 '18 at 22:27






  • 1




    @WesleyGroupshaveFeelingsToo Ah ok you are referring to that one $$ lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}$$ That's fine!
    – gimusi
    Nov 27 '18 at 22:28












  • In my book a friendly introduction to analysis by Kosmala, it is referred to 2.2.1C
    – Wesley Strik
    Nov 27 '18 at 22:28














  • 1




    That first proof looks good, and is how I would solve this problem.
    – Dave
    Nov 27 '18 at 22:22












  • Both ways are fine, but what do you mean with "the fraction limit theorem"?
    – gimusi
    Nov 27 '18 at 22:25










  • One of the standard results for limits is that if we take a limit of a fraction of two sequences, it is the same as the fraction of limits.
    – Wesley Strik
    Nov 27 '18 at 22:27






  • 1




    @WesleyGroupshaveFeelingsToo Ah ok you are referring to that one $$ lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}$$ That's fine!
    – gimusi
    Nov 27 '18 at 22:28












  • In my book a friendly introduction to analysis by Kosmala, it is referred to 2.2.1C
    – Wesley Strik
    Nov 27 '18 at 22:28








1




1




That first proof looks good, and is how I would solve this problem.
– Dave
Nov 27 '18 at 22:22






That first proof looks good, and is how I would solve this problem.
– Dave
Nov 27 '18 at 22:22














Both ways are fine, but what do you mean with "the fraction limit theorem"?
– gimusi
Nov 27 '18 at 22:25




Both ways are fine, but what do you mean with "the fraction limit theorem"?
– gimusi
Nov 27 '18 at 22:25












One of the standard results for limits is that if we take a limit of a fraction of two sequences, it is the same as the fraction of limits.
– Wesley Strik
Nov 27 '18 at 22:27




One of the standard results for limits is that if we take a limit of a fraction of two sequences, it is the same as the fraction of limits.
– Wesley Strik
Nov 27 '18 at 22:27




1




1




@WesleyGroupshaveFeelingsToo Ah ok you are referring to that one $$ lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}$$ That's fine!
– gimusi
Nov 27 '18 at 22:28






@WesleyGroupshaveFeelingsToo Ah ok you are referring to that one $$ lim_{k rightarrow infty} frac{1}{2+frac{1}{k}}$$ That's fine!
– gimusi
Nov 27 '18 at 22:28














In my book a friendly introduction to analysis by Kosmala, it is referred to 2.2.1C
– Wesley Strik
Nov 27 '18 at 22:28




In my book a friendly introduction to analysis by Kosmala, it is referred to 2.2.1C
– Wesley Strik
Nov 27 '18 at 22:28










2 Answers
2






active

oldest

votes


















1














Yes you are correct since $a_n not to 0$ the series does not converges and since it is with positive terms it diverges to $+infty$.



Also the alternative way is fine.






share|cite|improve this answer





















  • Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
    – Wesley Strik
    Nov 27 '18 at 22:37












  • in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
    – Wesley Strik
    Nov 27 '18 at 22:38












  • @WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
    – gimusi
    Nov 27 '18 at 22:42



















1














Or you can use the following $${kover 2k+1}ge {kover 3k}={1over 3}$$for $kge 1$ which implies a direct divergence. Still, your proof sounds right to me.






share|cite|improve this answer





















  • Too quick on getting rid of the initial $k$ haha.
    – Wesley Strik
    Nov 27 '18 at 22:39










  • Thanks, that was useful.
    – Wesley Strik
    Nov 27 '18 at 22:39






  • 1




    You're welcome. Good luck!
    – Mostafa Ayaz
    Nov 27 '18 at 22:40











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Yes you are correct since $a_n not to 0$ the series does not converges and since it is with positive terms it diverges to $+infty$.



Also the alternative way is fine.






share|cite|improve this answer





















  • Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
    – Wesley Strik
    Nov 27 '18 at 22:37












  • in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
    – Wesley Strik
    Nov 27 '18 at 22:38












  • @WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
    – gimusi
    Nov 27 '18 at 22:42
















1














Yes you are correct since $a_n not to 0$ the series does not converges and since it is with positive terms it diverges to $+infty$.



Also the alternative way is fine.






share|cite|improve this answer





















  • Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
    – Wesley Strik
    Nov 27 '18 at 22:37












  • in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
    – Wesley Strik
    Nov 27 '18 at 22:38












  • @WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
    – gimusi
    Nov 27 '18 at 22:42














1












1








1






Yes you are correct since $a_n not to 0$ the series does not converges and since it is with positive terms it diverges to $+infty$.



Also the alternative way is fine.






share|cite|improve this answer












Yes you are correct since $a_n not to 0$ the series does not converges and since it is with positive terms it diverges to $+infty$.



Also the alternative way is fine.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 22:23









gimusigimusi

1




1












  • Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
    – Wesley Strik
    Nov 27 '18 at 22:37












  • in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
    – Wesley Strik
    Nov 27 '18 at 22:38












  • @WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
    – gimusi
    Nov 27 '18 at 22:42


















  • Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
    – Wesley Strik
    Nov 27 '18 at 22:37












  • in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
    – Wesley Strik
    Nov 27 '18 at 22:38












  • @WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
    – gimusi
    Nov 27 '18 at 22:42
















Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
– Wesley Strik
Nov 27 '18 at 22:37






Is it fine because a sum is the limit of partial sums and for limits we know that a divergent limit to $infty$, times any positive constant still is a divergent limit?
– Wesley Strik
Nov 27 '18 at 22:37














in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
– Wesley Strik
Nov 27 '18 at 22:38






in maths: ${ s_n } rightarrow infty $ then for $c>0$, also ${ c s_n } rightarrow infty $
– Wesley Strik
Nov 27 '18 at 22:38














@WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
– gimusi
Nov 27 '18 at 22:42




@WesleyGroupshaveFeelingsToo Yes my observation is aimed to stress that form $a_n not to 0$ we can conclude in general that the series "does not converges" and in taht particular case it diverges to $infty$. I prefer use that distinction since we can also have the possibility of bounded but not convergent series.
– gimusi
Nov 27 '18 at 22:42











1














Or you can use the following $${kover 2k+1}ge {kover 3k}={1over 3}$$for $kge 1$ which implies a direct divergence. Still, your proof sounds right to me.






share|cite|improve this answer





















  • Too quick on getting rid of the initial $k$ haha.
    – Wesley Strik
    Nov 27 '18 at 22:39










  • Thanks, that was useful.
    – Wesley Strik
    Nov 27 '18 at 22:39






  • 1




    You're welcome. Good luck!
    – Mostafa Ayaz
    Nov 27 '18 at 22:40
















1














Or you can use the following $${kover 2k+1}ge {kover 3k}={1over 3}$$for $kge 1$ which implies a direct divergence. Still, your proof sounds right to me.






share|cite|improve this answer





















  • Too quick on getting rid of the initial $k$ haha.
    – Wesley Strik
    Nov 27 '18 at 22:39










  • Thanks, that was useful.
    – Wesley Strik
    Nov 27 '18 at 22:39






  • 1




    You're welcome. Good luck!
    – Mostafa Ayaz
    Nov 27 '18 at 22:40














1












1








1






Or you can use the following $${kover 2k+1}ge {kover 3k}={1over 3}$$for $kge 1$ which implies a direct divergence. Still, your proof sounds right to me.






share|cite|improve this answer












Or you can use the following $${kover 2k+1}ge {kover 3k}={1over 3}$$for $kge 1$ which implies a direct divergence. Still, your proof sounds right to me.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 '18 at 22:37









Mostafa AyazMostafa Ayaz

14.5k3937




14.5k3937












  • Too quick on getting rid of the initial $k$ haha.
    – Wesley Strik
    Nov 27 '18 at 22:39










  • Thanks, that was useful.
    – Wesley Strik
    Nov 27 '18 at 22:39






  • 1




    You're welcome. Good luck!
    – Mostafa Ayaz
    Nov 27 '18 at 22:40


















  • Too quick on getting rid of the initial $k$ haha.
    – Wesley Strik
    Nov 27 '18 at 22:39










  • Thanks, that was useful.
    – Wesley Strik
    Nov 27 '18 at 22:39






  • 1




    You're welcome. Good luck!
    – Mostafa Ayaz
    Nov 27 '18 at 22:40
















Too quick on getting rid of the initial $k$ haha.
– Wesley Strik
Nov 27 '18 at 22:39




Too quick on getting rid of the initial $k$ haha.
– Wesley Strik
Nov 27 '18 at 22:39












Thanks, that was useful.
– Wesley Strik
Nov 27 '18 at 22:39




Thanks, that was useful.
– Wesley Strik
Nov 27 '18 at 22:39




1




1




You're welcome. Good luck!
– Mostafa Ayaz
Nov 27 '18 at 22:40




You're welcome. Good luck!
– Mostafa Ayaz
Nov 27 '18 at 22:40


















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