# Does this limit involving the Dirichlet eta function and the Riemann zeta function make sense?

Let $p_n$ the sequence of prime numbers (and you will consider below, too, the sequence $\frac{1}{n}$ with $n>1$).

And if it isn't wrong for $0<\Re s<1$ the known equation between Dirichlet eta function and Riemann zeta function $$\eta(s)=(1-2^{1-s})\zeta(s).$$ I excuse the following question to refresh basics in mathematics (and try learn, when it is possible, the more easy facts and computations with these special functions). As is known and we can read in the first paragraphs of  Dirichlet eta function doesn't mask the zeros of Riemann zeta function, and $\eta(1)$ is neither infinite nor zero. I hope don't wrong in previous claim, but the exercise is only to know if my computations are feasible.

Question. It is possible to define and compute $$\lim_{n\to\infty}\frac{\frac{\eta(1-\frac{1}{p_n})}{\zeta(1-\frac{1}{p_n})}}{\frac{\eta(1-\frac{1}{n})}{\zeta(1-\frac{1}{n})}}?$$

My computations are evuluate previous quotients of complex numbers as the following real numbers, with $n>1$ $$\frac{1-2^{\frac{1}{p_n}}}{1-2^{\frac{1}{n}}},$$ as you see if this is right is it an exercise of calculus. After I evaluate this as $\frac{0}{0}$. I've used The Prime Number Theorem for write the asymptotic equivalence $p_n\sim n\log n$ for use L'Hôpital rule

$$\lim_{x\to\infty}\frac{1-2^{\frac{1}{x\log x}}}{1-2^{\frac{1}{x}}},$$ and conclude that this limit is $0$.

Can you give your computations to do an comparision and can you claim that the limit in previous Question has sense when we consider, now yes, the cited special functions? Thanks in advance.

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## 1 Answer

Both your result and your intuition are right. Let us formalize them. Assuming that $p_1 = 2$, we shall use the following classic upper bound: $p_n > n(\ln n + \ln \ln n - 1), \ n \ge 6$. This bound is good, but working with it would be annoying, computations becoming too long, therefore let us simplify it a bit: note that $n > {\rm e} ^{\rm e} \iff \ln \ln n - 1 > 0$, which gives us the inequality $p_n > n \ln n$ for $n > {\rm e} ^{\rm e}$. Despite being weaker, this inequality is much more comfortable to work with and, as you are going to see, enough for our needs.

The above inequality implies that

$$\frac {1 - 2 ^{\frac 1 {p_n}}} {1 - 2 ^{\frac 1 n}} < \frac {1 - 2 ^{\frac 1 {n \ln n}}} {1 - 2 ^{\frac 1 n}} ,$$

and through the change of variable $t = \frac 1 n$

$$\lim \limits _{n \to \infty} \frac {1 - 2 ^{\frac 1 {n \ln n}}} {1 - 2 ^{\frac 1 n}} = \lim \limits _{t \to 0, \ t>0} \frac {1 - 2 ^{\frac t {-\ln t}}} {1 - 2 ^t} .$$

Aplying l'Hôpital's theorem, the limit becomes

$$\lim \limits _{t \to 0, \ t>0} \frac {- 2 ^{\frac t {-\ln t}} \ln 2} {- 2 ^t \ln 2} \left( \frac t {-\ln t} \right)' .$$

Through the change of variable $t = \frac 1 u, \ u \to \infty$, the fraction $\dfrac {2 ^{\frac t {-\ln t}}} {2 ^t}$ can be rewritten as $\dfrac {2 ^{\frac 1 {u \ln u}}} {2 ^{\frac 1 u}}$ which obviously tends to $\dfrac {2^0} {2^0} = 1$, so we only have to concentrate on the limit of the derivative:

$$\lim \limits _{t \to 0, \ t>0} \left( \frac t {-\ln t} \right)' = \lim \limits _{t \to 0, \ t>0} \frac {-\ln t + 1} {(-\ln t)^2} .$$

The change of variable $u = -\ln t, \ u \to \infty$ transforms this into

$$\lim \limits _{u \to \infty} \frac {u + 1} {u^2} ,$$

which is obviously $0$. Multiplying the limits of the two subexpressions ($1$ and $0$), we get that

$$\lim \limits _{n \to \infty} \frac {1 - 2 ^{\frac 1 {n \ln n}}} {1 - 2 ^{\frac 1 n}} = 1 \cdot 0 = 0 .$$

On the other hand, $0 \le \dfrac {1 - 2 ^{\frac 1 {p_n}}} {1 - 2 ^{\frac 1 n}}$ because both the numerator and the denominator are negative.

Collecting all the results obtained so far, we may finally write

$$0 \le \lim \limits _{n \to \infty} \dfrac {1 - 2 ^{\frac 1 {p_n}}} {1 - 2 ^{\frac 1 n}} \le \lim \limits _{n \to \infty} \frac {1 - 2 ^{\frac 1 {n \ln n}}} {1 - 2 ^{\frac 1 n}} = 0 ,$$

and this is a rigorous proof of the fact that $$\lim \limits _{n \to \infty} \dfrac {1 - 2 ^{\frac 1 {p_n}}} {1 - 2 ^{\frac 1 n}} = 0 .$$