Dirichlet series

Prove: if the Dirichlet series $\displaystyle \sum_{n=1}^{\infty } \frac{a_{n}}{n^{x}}$ converges at $x=x_{0}$, then this series also converges when $x>x_{0}$.

Prove that if the series $\displaystyle \sum_{n=1}^{\infty } a_{n}$ converges, then the Dirichlet series $\displaystyle \sum_{n=1}^{\infty } \dfrac {a_{n}}{n^{x}}$ converges uniformly for $x \geq 0$.

Compute: $\displaystyle \lim_{x \rightarrow +0} \sum_{n=1}^{\infty } \frac{1}{2^{n}n^{x}}$.

Prove the theorem: if a general Dirichlet series converges at $\sigma = \sigma_{0}$, then it converges for any $s$ for which $\sigma > \sigma_{0}$.

Prove the theorem: if for the quantity $A_{n} = \sum_{\nu =1}^{n} a_{\nu }$ the identity $A_{n} = O(n^{\sigma_{0}})$, $\sigma_{0} \geqslant 0$, holds, then the Dirichlet series $\sum \dfrac {a_{n}}{n^{s}}$ converges for $\sigma > \sigma_{0}$, and converges absolutely and uniformly for $\sigma > \sigma_{0}+1+\delta$, where $\delta$ is any positive constant.

Prove the theorem: if the series $\sum_{n=1}^{\infty } \dfrac {a_{n}}{n^{s}}$ converges at $\sigma = \sigma_{0} \geqslant 0$, then $A_{n} = O(n^{\sigma_{0}})$.

Using Pringsheim's theorem, prove that the abscissa of convergence of the general Dirichlet series $\sum p_{n}^{-s}$ is $\overline{\lim }\dfrac {\ln n}{\ln p_{n}}$.

Let the polynomials $P_1(x), P_2(x), \ldots , P_l(x)$ be $\not\equiv 0$ and of degree $m_1-1, m_2-1, \ldots , m_l-1$ respectively, and let the real constants $a_1, a_2, \ldots , a_l$ be distinct. The function

has at most $m_1+m_2+\cdots +m_l-1$ real zeros.

Let $a_1, a_2, \ldots , a_n, \lambda_1, \lambda_2, \ldots , \lambda_n$ be real constants, $\lambda_1 < \lambda_2 < \lambda_3 < \cdots < \lambda_n$. Denote by $Z$ the number of real zeros of the entire function

and by $C$ the number of changes of sign in the sequence of numbers $a_1, a_2, \ldots , a_n$. Then $C-Z$ is a non-negative even integer. (A generalization of 36, 37 different from 41: Replace $e^x$ by $x$.)

Proof. We may assume without loss of generality that $a_1, a_2, \ldots a_n$ are all different from $0$. That $C-Z$ is even is evident from the fact that as $x \to -\infty$ the term $a_1 e^{\lambda_1 x}$ is the dominant term, and as $x \to +\infty$ the term $a_n e^{\lambda_n x}$ is the dominant term. [8, 9, 37.] That $C-Z \geq 0$ may be proved by mathematical induction; we deduce the relation for $C$ from that for $C-1$, with the aid of Rolle's theorem [solution 75]. Indeed, if there are no changes of sign, evidently $Z$ is also $=0$ and the statement holds. Assume that it holds in the case that there are $C-1$ changes of sign. $F(x)$ has $C$ changes of sign, $C \geq 1$; e.g. let $\alpha +1$ be an index of change, $1 \leq \alpha < n$, $a_{\alpha }a_{\alpha +1} < 0$. Choose a number $\lambda$ in the interval $\lambda_{\alpha } < \lambda < \lambda_{\alpha +1}$ and consider the function

Let the number of zeros of $F^*(x)$ be $Z^*$. We have [6, 12]

Let the number of changes of sign of the sequence of coefficients

be $C^*$. Clearly

By the assumption of the mathematical induction argument we have

From the three relations above it follows that

q.e.d.---Was it necessary to assume from the outset that $a_1, a_2, \ldots , a_n$ are all $\neq 0$? Could we not also have chosen $\lambda =\lambda_{\alpha }$ or $\lambda =\lambda_{\alpha +1}$?

Let $\lambda_1 < \lambda_2 < \lambda_3 < \cdots$, $\lim_{n \to \infty } \lambda_n = \infty$. In the interior of its region of convergence (a half-plane bounded on the left), the Dirichlet series

has not more real zeros than the number of changes of sign of the sequence of coefficients $a_1, a_2, a_3, \ldots , a_n, \ldots$. (Generalization of 38.)

The number of positive zeros lying in the interior of the region of convergence does not exceed the number of changes of sign of the sequence

(Generalization of 44.) [80.]

The entire function

has simple zeros at the points

and no other real zeros. (Interpret $k^x$ as $e^{x \log k}$, with the principal value of $\log k$.)

Let $C$ be the number of changes of sign of the finite sequence

$\lambda_i$ ($i=1,2,\ldots ,n$) real numbers satisfying

and

where we assume that $D(0)=0$. Then $D(x)$ has at most $C+1$ real zeros.

(Both the hypothesis and the conclusion are stronger than in 83. Generalization of 44.1.)