# Convergence of a series

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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HallsofIvy
Homework Helper
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.

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??

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)$$\sum =\frac{2^1}{1}+\frac{2^2}{1+\frac{1}{2}}+\frac{2^3}{1+\frac{1}{2}+\frac{1}{3}}+...$$

OR perhaps

b)$$\sum =2^{\frac{1}{1}}+2^{\frac{2}{1+1/2}}+2^{\frac{3}{1+1/2+1/3}}+...$$

Something else?

Anycase,both are divergent becouse :

$$\lim_{n\to \infty}a_{n}>0$$

Also consider the ratio test: if $$a_(n+1)/(a_n) >1,$$ the series diverges; $$if a_(n+1)/a_n =1,$$result unknow; if $$a_(n+1)/a_n<1$$ the series converges.

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Also consider the ratio test: if $$a_(n+1)/(a_n) >1,$$ the series diverges; $$if a_(n+1)/a_n =1,$$result unknow; if $$a_(n+1)/a_n<1$$ 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.

Do you mean:

a)$$\sum =\frac{2^1}{1}+\frac{2^2}{1+\frac{1}{2}}+\frac{2^3}{1+\frac{1}{2}+\frac{1}{3}}+...$$

OR perhaps

b)$$\sum =2^{\frac{1}{1}}+2^{\frac{2}{1+1/2}}+2^{\frac{3}{1+1/2+1/3}}+...$$

Something else?

Anycase,both are divergent becouse :

$$\lim_{n\to \infty}a_{n}>0$$

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???

Gib Z
Homework Helper
Ok well we know that the harmonic series
$$\sum_{n=0}^{\infty} \frac{1}{n}$$ 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.

Do you mean:

a)$$\sum =\frac{2^1}{1}+\frac{2^2}{1+\frac{1}{2}}+\frac{2^3}{1+\frac{1}{2}+\frac{1}{3}}+...$$

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

a_n=???

Ok well we know that the harmonic series
$$\sum_{n=0}^{\infty} \frac{1}{n}$$ 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??

Notation?

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

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

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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Gib Z
Homework Helper
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??

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 $$2^\frac{n}{\sum_k^{\infty} \frac{1}{k}}$$. 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.

Gib Z
Homework Helper
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
$$\frac{2^3}{1 +1/2+1/3}$$ Is $$\frac{2^3}{1+1+1}$$. In other words, divide by the larger series to get a smaller term by term. Since we can see that $$\frac{2^n}{n}$$ when n approaches infinity is obviously more than zero, so it doesn't converge.

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
$$\frac{2^3}{1 +1/2+1/3}$$ Is $$\frac{2^3}{1+1+1}$$. In other words, divide by the larger series to get a smaller term by term. Since we can see that $$\frac{2^n}{n}$$ 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.

Gib Z
Homework Helper
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.

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