2
$\begingroup$

What's the result of this two sequences ?

$\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{n+C}$ $\frac{\log(n)}{n} + \frac{\log(n+1)}{n+1}+\frac{\log(n+2)}{n+2} + ... + \frac{\log(n+C)}{n+C}$

EDIT: I mean $\sum_{i=1}^{C} \frac{1}{n+i}$ and $\sum_{i=1}^{C} \frac{log(n+i)}{n+i}$, and yes C is a constant. And what I want is expressions for this sums, whithout iterating over i.

  • 0
    Assuming $C$ is constant, the bounds for first sum $S$ is $\frac{C}{n} \geq S \geq \frac{2C}{2n+C+1}$, and therefore approaches $0$ as $n \to \infty$2012-03-20

1 Answers 1

2

The first sum can be written as $\Psi(n+C+1) - \Psi(n)$, the second as $\lim_{z \to 1+} \left(\zeta(1,z,n) - \zeta(1,z,n+C+1)\right)$ where $\zeta(1,z,n)$ is the derivative (with respect to $z$) of the Hurwitz zeta function. Those expressions may not help you much, if you look at the definitions of those functions. But they may help with asymptotics, e.g. according to Maple, as $n \to \infty$ with $C$ fixed, $ \Psi(n+C+1) - \Psi(n) = {\frac {C+1}{n}}-{\frac {C \left( C+1 \right) }{2 {n}^{2}}}+{ \frac {C \left( C+1 \right) \left( 2\,C+1 \right) }{6{n}^{3}}}-{ \frac {{C}^{2} \left( C+1 \right) ^{2}}{4{n}^{4}}}+{\frac {C \left( C+1 \right) \left( 2\,C+1 \right) \left( 3\,{C}^{2}+3\,C-1 \right) }{30 {n}^{5}}}+O \left( {n}^{-6} \right) $