Convergence of a series

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 a_n goes to infinity and, therefore, the series (the sum), if that was what you meant, does not converge.

 

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]

 

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.

 

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.

 

Notation?

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.

 

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.

 

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.

 

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.