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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.

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By result you mean limit as $n \to \infty$? I presume $C$ is a constant? – Aryabhata Mar 20 '12 at 19:04
There is no simple expression for these sums, if that is what you are asking for. But there are useful estimates. – André Nicolas Mar 20 '12 at 19:05
@user995434: We can see it is a sum. The question is: What exactly do you mean by "result"? – Aryabhata Mar 20 '12 at 19:07
@Aryabhata I mean an expression for this sum, whithout iterating over i. – shn Mar 20 '12 at 19:10
@AndréNicolas what are the estimates then ? – shn Mar 20 '12 at 19:17
up vote 2 down vote accepted

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, $$ \Psi(n+C+1) - \Psi(n) = {\frac {C+1}{n}}-{\frac {C \left( C+1 \right) }{2 {n}^{2}}}+{ \frac {C \left( C+1 \right) \left( 2\,C+1 \right) }{6{n}^{3}}}-{ \frac {{C}^{2} \left( C+1 \right) ^{2}}{4{n}^{4}}}+{\frac {C \left( C+1 \right) \left( 2\,C+1 \right) \left( 3\,{C}^{2}+3\,C-1 \right) }{30 {n}^{5}}}+O \left( {n}^{-6} \right) $$

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