Riemann Zeta approaches infinity as x approaches 1

In summary, the given conversation discusses the proof that the sum of the functional series \sum_{n=1}^{\infty}\frac{1}{n^x} approaches infinity as x approaches 1 from the right. This is proven by showing that as x approaches 1, the series approaches the harmonic series, which diverges. The formal proof involves using the fact that the harmonic series diverges and manipulating the given functional series to show that it also diverges.
  • #1
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1
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Homework Statement




We know [itex]\sum_{n=1}^{\infty}\frac{1}{n^x}[/itex] is uniformly convergent on the interval [itex]x\in(1,\infty)[/itex] and that its sum is called [itex]\zeta(x)[/itex]. Proof that [itex]\zeta(x) \rightarrow \infty[/itex] as [itex]x \rightarrow 1^+[/itex].


Homework Equations



We cannot find the formula that [itex]\zeta[/itex] is given by, but we know
[itex]\sum_{n=1}^{\infty}\frac{1}{n^1}[/itex] diverges.

The Attempt at a Solution


I've already shown that the functional series is uniformly convergent on the above interval and that the derivative also converges uniformly to a limit function, thus [itex]\zeta(x)[/itex] is continuous and can be differentiated.

The solution seem quite apparent: As [itex]x[/itex] approaches [itex]1[/itex], the series will approach the harmonic series, and thus it will diverge. Given any [itex]M>0[/itex], I can find a [itex]b>0[/itex] so that [itex]x \in (1,1+b) \Rightarrow \zeta(x) >M[/itex], as we know that we will approach the harmonic series.

However, I'm not quite sure how I can go about formulating a formal proof. For functional series, I suppose I cannot just plug in the limit and say its a harmonic series (especially as it's not defined there).

Under which conditions can I be completely sure that plugging in the relevant [itex]x[/itex] and then let [itex]n[/itex] approach infinity will be equivalent to letting n approach infinity and then let [itex]x[/itex] approach infinity? Or can I approach this in a different way?

Any help will be greatly appreciated!
 
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  • #2
Why not use the fact that
$$
\sum_{n=1}^{\infty}\frac{1}{n}
$$
is a harmonic series.

Then
$$
1+\frac{k}{2}\leq\sum_{n=1}^{2^k}\frac{1}{n}
$$
as ##k\to\infty##
$$
\infty = 1+\frac{k}{2}\leq \sum_{n=1}^{2^k}\frac{1}{n}
$$
 

1. What is the Riemann Zeta function?

The Riemann Zeta function, denoted as ζ(x), is a mathematical function that is defined for all complex numbers except x = 1. It is closely related to the distribution of prime numbers and is an important tool in number theory and analytic number theory.

2. What does it mean for the Riemann Zeta function to approach infinity?

When we say that the Riemann Zeta function approaches infinity, it means that as the variable x gets closer and closer to 1, the value of the function also gets larger and larger without bound. In other words, the function grows without any limit.

3. How does the Riemann Zeta function approach infinity as x approaches 1?

The Riemann Zeta function approaches infinity as x approaches 1 in a logarithmic fashion. This means that the rate of increase of the function slows down as it approaches infinity, and it never actually reaches infinity at x = 1.

4. Why is it significant that the Riemann Zeta function approaches infinity as x approaches 1?

The fact that the Riemann Zeta function approaches infinity as x approaches 1 is significant because it is closely related to the Riemann Hypothesis, one of the most famous unsolved problems in mathematics. This behavior of the function is also important in understanding the distribution of prime numbers and has many applications in number theory.

5. Is there a specific value for the Riemann Zeta function at x = 1?

No, there is not a specific value for the Riemann Zeta function at x = 1. As mentioned earlier, the function grows without bound as x approaches 1, and it is undefined at that point. However, there are methods for assigning a value to the function at x = 1, such as using analytic continuation or the functional equation of the Riemann Zeta function.

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