Does the Series \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right) Converge?

[Sweet_GirL]

hello
I have this one:

\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)


mmmmm am sure it will be tested by using one of the comparison tests
but am not getting it
any help?

this is not my homework, actually I finished my college 2 years ago.

 

[rs1n]

I would start by looking at the behavior of just \sqrt[n]{n} as n\to\infty. Use the techniques of logarithms to find \lim \sqrt[n]{n}. Once you see what \sqrt[n]{n} approaches, you should immediately be able to see how

<br /> \sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right) <br />

must behave. If you don't see it, just consider what (1-\sqrt[n]{n}) must approach, knowing the limit of \sqrt[n]{n}.

 

[Sweet_GirL]

well,
1 - \sqrt[n]{n} \rightarrow 0 as n \rightarrow \infty

and this will make no sense.

 

[rs1n]

well,
1 - \sqrt[n]{n} \rightarrow 0 as n \rightarrow \infty

and this will make no sense.

You are quite right; in my zeal, I made a mistake in computing the limit of the n-th root of n and getting 0!

 

[l'Hôpital]

I might be wrong but try Cauchy's Condensation Test.
<br /> 1 - n^{\frac{1}{n}} = 2^k (1 - 2^{\frac{k}{2^k}})<br />

Which obviously fails the limit test...

 

[Sweet_GirL]

It must be solved by the standart test.

Anyone ?

 

[rs1n]

In order to use the CCT your terms need to be positive and non-increasing. This isn't a big deal since we can just negate the sum, and consider the sum starting from n=3.

As for using "standard" tests, what about the integral test? I've only thought as far as:

\int_3^\infty (\sqrt[x]{x}-1)\ dx \ge \int_3^\infty (\sqrt[x]{3}-1)\ dx = \int_3^\infty (3^{1/x}-1)\ dx

 

[Gib Z]

Theorem: A bounded monotonic sequence converges.

 

[rs1n]

Theorem: A bounded monotonic sequence converges.

Yes, but a sequence is very different from a series. Unless you are referring to the partial sums; but this would require that you can bound the partial sums. Can you elaborate on how this is done?

 

[Gib Z]

The n-th root of n is greater than the n-th root of 1.

 

[Sweet_GirL]

Still searching for a solution with standard tests ..

 

[Bohrok]

Try finding a function f(x) such that \sqrt[x]{x} - 1 >> f(x) using l'Hôpital's rule, \sqrt[x]{x} - 1 > f(x) on [1, ∞), and \sum_{n=1}^\infty f(n) diverges. Then use the comparison test on your series and f(n).

 

[Sweet_GirL]

I tried that
but its not easy to find that f
and also f must be positive

 

[g_edgar]

n^{1/n}-1 is asymptotic to \log(n)/n, so you need to analyze the convergence of \sum\log(n)/n

 

[rs1n]

n^{1/n}-1 is asymptotic to \log(n)/n, so you need to analyze the convergence of \sum\log(n)/n

Ahh, this indeed will do it. Very clever.