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Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$)

((This is to be done by writing $L(s)=exp(\sum_{p} \frac{a_{p}}{p^s})G(s)$, where $G(s)$ is a harmless function and by partial summation i.e., $\sum_{p \leq x} a_{p} \ll x^{1-\epsilon}$ )) .
Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.

Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$)

((This is to be done by writing $L(s)=exp(\sum_{p} \frac{a_{p}}{p^s})G(s)$, where $G(s)$ is a harmless function and by partial summation together with the bound, $\sum_{p \leq x} a_{p} \ll x^{1-\epsilon}$ )) .
Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.

Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$)

((This is to be done by writing $L(s)=exp(\sum_{p} \frac{a_{p}}{p^s})G(s)$, where $G(s)$ is a harmless function and by partial summation i.e $\sum_{p \leq x} a_{p} \ll x^{1-\epsilon}$ )) .
Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.

Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$)

((This is to be done by writing $L(s)=exp(\sum_{p} \frac{a_{p}}{p^s})G(s)$, where $G(s)$ is a harmless function and by partial summation i.e., $\sum_{p \leq x} a_{p} \ll x^{1-\epsilon}$ )) .
Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.

Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n=1} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$). Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.

Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$)

((This is to be done by writing $L(s)=exp(\sum_{p} \frac{a_{p}}{p^s})G(s)$, where $G(s)$ is a harmless function and by partial summation i.e $\sum_{p \leq x} a_{p} \ll x^{1-\epsilon}$ )) .
Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.