Convergence of a series

  • #1
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convergence of a series. NEED HELP!!!!!!

i am sorry for my symbols, but i cant use latex yet.

i have this series a_n=2^n/(1+1/2+1/3+1/4+.....+1/n), and i am asked to find the convergence of this series when n goes from 1 (n=1) to infinity.

I have done this series, however i am not sure if i can go here, i do not know if i can rewrite the given series in this form, and not to change anything, so it would be:

a_n=2^n/(1/n), when n goes from 1 to infinity. If i am right here, then the rest is all clear to me.

any help would do. SO what do you guys think???
 
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Answers and Replies

  • #2
HallsofIvy
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i am sorry for my symbols, but i cant use latex yet.

i have this series a_n=2^n/(1+1/2+1/3+1/4+.....+1/n), and i am asked to find the convergence of this series when n goes from 1 (n=1) to infinity.

I have done this series, however i am not sure if i can go here, i do not know if i can rewrite the given series in this form, and not to change anything, so it would be:

a_n=2^n/(1/n), when n goes from 1 to infinity. If i am right here, then the rest is all clear to me.

any help would do. SO what do you guys think???
I don't see how you can possible think that 2n/(1/n)= n2n is the same as what you give above. In any case, it seems clear to me, since 2n increases much faster than n while 1+ 1/2+ 1/3+ ...+ 1/n increases slower than n, that the sequence an goes to infinity and, therefore, the series (the sum), if that was what you meant, does not converge.
 
  • #3
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I don't see how you can possible think that 2n/(1/n)= n2n is the same as what you give above. In any case, it seems clear to me, since 2n increases much faster than n while 1+ 1/2+ 1/3+ ...+ 1/n increases slower than n, that the sequence an goes to infinity and, therefore, the series (the sum), if that was what you meant, does not converge.
yeah i also think it cannot converge, but how can i show it then?
I mean how can i come to that conclusion in a more elaborated way??
can you give me some hints then? Because look here what would be the partial sum for example lets say when we take n=1, then n=2, it looks a little wierd to me writin let's say when we take n=1

S_1=2^1/(1+1/2+1/3+.....+1/1)

it looks a little wierd doesn't it??
 
  • #4
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0
i am sorry for my symbols, but i cant use latex yet.

i have this series a_n=2^n/(1+1/2+1/3+1/4+.....+1/n), and i am asked to find the convergence of this series when n goes from 1 (n=1) to infinity.

I have done this series, however i am not sure if i can go here, i do not know if i can rewrite the given series in this form, and not to change anything, so it would be:

a_n=2^n/(1/n), when n goes from 1 to infinity. If i am right here, then the rest is all clear to me.

any help would do. SO what do you guys think???
Do you mean:

a)[tex]\sum =\frac{2^1}{1}+\frac{2^2}{1+\frac{1}{2}}+\frac{2^3}{1+\frac{1}{2}+\frac{1}{3}}+...[/tex]

OR perhaps

b)[tex]\sum =2^{\frac{1}{1}}+2^{\frac{2}{1+1/2}}+2^{\frac{3}{1+1/2+1/3}}+...[/tex]

Something else?

Anycase,both are divergent becouse :

[tex]\lim_{n\to \infty}a_{n}>0[/tex]
 
  • #5
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Also consider the ratio test: if [tex]a_(n+1)/(a_n) >1,[/tex] the series diverges; [tex]if a_(n+1)/a_n =1, [/tex]result unknow; if [tex] a_(n+1)/a_n<1[/tex] the series converges.
 
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  • #6
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Also consider the ratio test: if [tex]a_(n+1)/(a_n) >1,[/tex] the series diverges; [tex]if a_(n+1)/a_n =1, [/tex]result unknow; if [tex] a_(n+1)/a_n<1[/tex] the series converges.

yeah i have tried this. It is also called dallamber's rule of convergence. but the way it is given as i OP-ed, i cannot really simplify, or at least have no idea how to simplify.
 
  • #7
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Do you mean:

a)[tex]\sum =\frac{2^1}{1}+\frac{2^2}{1+\frac{1}{2}}+\frac{2^3}{1+\frac{1}{2}+\frac{1}{3}}+...[/tex]

OR perhaps

b)[tex]\sum =2^{\frac{1}{1}}+2^{\frac{2}{1+1/2}}+2^{\frac{3}{1+1/2+1/3}}+...[/tex]

Something else?

Anycase,both are divergent becouse :

[tex]\lim_{n\to \infty}a_{n}>0[/tex]
i took that serie as it it originally written in the book. the proffesor put it on an exam once.
so it is a_n=2^n/(1+1/2+1/3+....+1/n), when n goes from 1 to infinity.

the thing that appears to be a prob to me, and that looks very wierd, is the denominator. because how would you rewrite this when n=1????????, so i mean the partial sum of this series when n=1???????

can anyone else give some more hints???
 
  • #8
Gib Z
Homework Helper
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Ok well we know that the harmonic series
[tex]\sum_{n=0}^{\infty} \frac{1}{n}[/tex] diverges. Look on wikipedia for a proof. Since that sum diverges to positive infinity, n over that sum is 0. 2^0 is 1, larger than zero. Does not converge. A must longer and stupider way than tehno's.
 
  • #9
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Do you mean:

a)[tex]\sum =\frac{2^1}{1}+\frac{2^2}{1+\frac{1}{2}}+\frac{2^3}{1+\frac{1}{2}+\frac{1}{3}}+...[/tex]
how would you write the general term for this series
i mean

a_n=???
 
  • #10
1,631
4
Ok well we know that the harmonic series
[tex]\sum_{n=0}^{\infty} \frac{1}{n}[/tex] diverges. Look on wikipedia for a proof. Since that sum diverges to positive infinity, n over that sum is 0. 2^0 is 1, larger than zero. Does not converge. A must longer and stupider way than tehno's.
yeah i know the proof for 1/n .
but here n goes from 1 to infinity not from 0.
Moreover, i do know that this series does not converge, however what i am lookin for is a way, an elaborated way, to come to that conclusion.
In any case i am not following you here Gib Z. Can u give more explanations, maybe it will be boring for u but i just dont get it!!
Stupid, no??
 
  • #11
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Notation?

how would you write the general term for this series i mean

a_n=???
sutupidmath ,that infinite sum can be formally written as:
[tex]\sum_{n=1}^{\infty}\frac{2^n}{\sum_{i=1}^{n}(1/i)}[/tex]

Did you denoted by a_n ,n-th partial sum ?.Usually partial sum is denoted by s_n.Please try by yourself ,after this, to see the notation for the partial sum.
However,I think that the notation is the least important here.
 
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  • #12
Gib Z
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yeah i know the proof for 1/n .
but here n goes from 1 to infinity not from 0.
Moreover, i do know that this series does not converge, however what i am lookin for is a way, an elaborated way, to come to that conclusion.
In any case i am not following you here Gib Z. Can u give more explanations, maybe it will be boring for u but i just dont get it!!
Stupid, no??
Really, tehno already answered your question, but here's my long way:

A series can not converge if the "last" term is more than zero. I think you already know that. So to prove that this series you have doesn't converge, we can see that in the final term, there will be [tex]2^\frac{n}{\sum_k^{\infty} \frac{1}{k}}[/tex]. The denomiator doesn't converge, and diverges to infinity.
n divided by infinty is 0. 2^0 is 1. The final term is not 0, it is 1.

It doesn't converge.
 
  • #13
Gib Z
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Actually I just realised my response was to tehno's shown form of b and you wanted A.

My answer is compare it to the smaller series Im giving here, and show that diverges.

When we divide the numerator of the term by a larger number, the term is overall smaller. So eg for n=3, a smaller term than
[tex]\frac{2^3}{1 +1/2+1/3}[/tex] Is [tex]\frac{2^3}{1+1+1}[/tex]. In other words, divide by the larger series to get a smaller term by term. Since we can see that [tex]\frac{2^n}{n}[/tex] when n approaches infinity is obviously more than zero, so it doesn't converge.
 
  • #14
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Actually I just realised my response was to tehno's shown form of b and you wanted A.

My answer is compare it to the smaller series Im giving here, and show that diverges.

When we divide the numerator of the term by a larger number, the term is overall smaller. So eg for n=3, a smaller term than
[tex]\frac{2^3}{1 +1/2+1/3}[/tex] Is [tex]\frac{2^3}{1+1+1}[/tex]. In other words, divide by the larger series to get a smaller term by term. Since we can see that [tex]\frac{2^n}{n}[/tex] when n approaches infinity is obviously more than zero, so it doesn't converge.
thank you all guys, in particular you Gib Z, because this is exactly what i was looking for. thnx again.
 
  • #15
Gib Z
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No problem, but next time if you really want to learn instead of having things pointed out to you, go with tehno's advice, it was really good advice.
 
  • #16
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No problem, but next time if you really want to learn instead of having things pointed out to you, go with tehno's advice, it was really good advice.
yeah, i really do appreciate tehno's advice and help, but i got a little confused so i needed some more hints. However i will give more effort other times, in solving problems by myself. thnx again
 

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