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
    By result you mean limit as $n \to \infty$? I presume $C$ is a constant?2012-03-20
  • 1
    There is no simple expression for these sums, if that is what you are asking for. But there are useful estimates.2012-03-20
  • 0
    @user995434: We can see it is a sum. The question is: What exactly do you mean by "result"?2012-03-20
  • 0
    @Aryabhata I mean an expression for this sum, whithout iterating over i.2012-03-20
  • 0
    @AndréNicolas what are the estimates then ?2012-03-20
  • 0
    @user995434: recall that $\int_1^x dt/t = \log x$ and consider Riemann sums.2012-03-20
  • 0
    Many estimates. The second harder sum is roughly $\int_n^{n+C}\frac{\log t}{t}\,dt$. And an antiderivative of our function is $(1/2)\log^2 t$. My "roughly" can be replaced by explicit upper and lower bounds. The first sum is similar, except the integration is easier. For the first, you might want to look up Euler's number $\gamma$.2012-03-20
  • 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, $$ \Psi(n+C+1) - \Psi(n) = {\frac {C+1}{n}}-{\frac {C \left( C+1 \right) }{2 {n}^{2}}}+{ \frac {C \left( C+1 \right) \left( 2\,C+1 \right) }{6{n}^{3}}}-{ \frac {{C}^{2} \left( C+1 \right) ^{2}}{4{n}^{4}}}+{\frac {C \left( C+1 \right) \left( 2\,C+1 \right) \left( 3\,{C}^{2}+3\,C-1 \right) }{30 {n}^{5}}}+O \left( {n}^{-6} \right) $$