Does the summation of 2^1/n -1 from n=1 to infinity converge or diverge?

Answer 1

Diverges. I'll give the proof later, in a hurry now.

OK, now let's prove it diverges. I'll prove for the general case Sum (n= 1, oo) (a^(1/n) -1), where a > 1. I assume you mean Sum (n =1, oo) (2^(1/n) - 1).

According to the properties of the exponential function, if a> 0, the a^x >= 1 + x ln(a) for every real x, with equality if, and only if. x = 0. So, if we replace x by 1/x, with x<>), we see that

a^(1/x) > 1 + ln(a)/x for every x <> 0. If a >1, then ln(a) > 0 and, for every integer n =1, 2,3...., we have

a^(1/n) > 1 + ln(a)/n => a^(1/n) -1 > ln(a)/n (1).

We know the harmonic series Sum(n =1, oo) 1/n diverges to oo. Since ln(a) >0, Sum(n =1, oo) ln(a)/n also goes to oo. So, by comparison, it follows from (1) that Sum(n= 1, oo) (a^(1/n) -1) --> oo, that is, diverges.

In your case, putting a =2 >1, we conclude the series diverges.

I'd like to point out that a^(1/n) -1 --> 0 when n--> oo, because 1/n -> 0. But the series diverges anyway.

Answer 2

Diverges by means of the nth term divergence try. This try states that if the shrink of a chain isn't equivalent to 0, then the sequence diverges. The shrink as n strategies infinity of (n+a million)/(2n-3) is a million/2. bear in mind, while taking limits with n drawing close infinity, if the the optimal exponent of the numerator equals that of the denominator, the shrink is only the ratio of the coefficients. simply by fact the stages in this problem are a million on the perfect and backside, the shrink is a million:2 or a million/2. simply by fact that a million/2 does not equivalent 0 the sequence diverges.

Answer 3

diverges. The nth term does not go to 0 and for a series to converge, the nth term has to go to zero.
EDIT: If you meant 2^(1/n) - 1 ( which is really how i should be viewed but I have grown accustomed to yahoo users not using brackets when they are supposed to) view Steiner's answer.