- #1
Klaus_Hoffmann
- 85
- 1
i don't know where i saw the question is does this expression (Cesaro sum ??) make sense for big lambda and positive delta
\sum_{n \le \lambda } (1- \frac{n}{\lambda})^{\delta} \mu (n)=S( \lambda , \delta)
for finite lambda the S is finite the question is what would happen as lambda goes to infinity ?, does the sum converge in the sense that for lambda big.
S \sim A+ O( \lambda ^{-b} (b a positive real number)
and the same for
\sum_{n \le \lambda } (1- \frac{n}{\lambda})^{\delta} \Lambda (n)=S( \lambda , \delta)
\sum_{n \le \lambda } (1- \frac{n}{\lambda})^{\delta} \mu (n)=S( \lambda , \delta)
for finite lambda the S is finite the question is what would happen as lambda goes to infinity ?, does the sum converge in the sense that for lambda big.
S \sim A+ O( \lambda ^{-b} (b a positive real number)
and the same for
\sum_{n \le \lambda } (1- \frac{n}{\lambda})^{\delta} \Lambda (n)=S( \lambda , \delta)
