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Let $0 \leq a \leq 1$ be a real number. I would like to know how to prove that the following sequence converges:

$$u_n(a)=\sum_{k=1}^n k^a- n^a \left(\frac{n}{1+a}+\frac{1}{2}\right)$$

For $a=1$:

$$u_n(1)=\sum\limits_{k=1}^{n} k- n \left(\frac{n}{1+1}+\frac{1}{2}\right)= \frac{n(n+1)}{2}-\frac{n(n+1)}{2}=0$$

so $u_n(1)$ converges to $0$.

for $a=0$: $$u_n(0)=\sum\limits_{k=1}^{n} 1- \left(\frac{n}{1+0}+\frac{1}{2}\right) = n-n+\frac{1}{2}=\frac{1}{2}$$

so $u_n(0)$ converges to $1/2$.

In genaral, the only idea I have in mind is the Cauchy integral criterion but it does not work because $k^a$ is an increasing function,

Do the proof involves Zeta serie ?

lara
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  • I am not sure what your series is...Could you be clearer? – Vishal Gupta Jul 06 '13 at 15:18
  • It's not a serie that why I wrote generalized Euler constant but you edited my title. It's a sequence, the sequence $u_n(a)$ which depends on $a$. – lara Jul 06 '13 at 15:25
  • @lara Have you tried the test Kummer? See it in http://140.177.205.23/KummersTest.html – Elias Costa Jul 06 '13 at 15:30
  • If you look at the proof for the Cauchy integral criterion, you'll see that if $f$ is increasing you can also compare $\sum f(n)$ with $\int f(t) , dt$. This allows you to compare $\sum_{k = 1}^n k^a$ with $\frac{n^{a+1}}{a+1}$ – Joel Cohen Jul 06 '13 at 15:31

2 Answers2

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To avoid a black box formula

Since the function $x\mapsto x^a$ is increasing then we find easily the asymptotic expansion

$$\sum_{k=1}^n k^a\sim_\infty \int_1^n x^a dx\sim_\infty \frac{1}{a+1}n^{a+1}\tag{AE}$$ hence we have $$\sum_{k=1}^n k^a=\frac{1}{a+1}n^{a+1}+O\left(n^{a}\right)\tag{1}$$ Now let's improve the equality $(1)$ so let $$v_n=\sum_{k=1}^n k^a-\frac{1}{a+1}n^{a+1}$$ so we have $$v_{n+1}-v_n=(n+1)^a-\frac{1}{a+1}(n+1)^{a+1}+\frac{1}{a+1}n^{a+1}$$ then we expand using the binomial formula we find $$v_{n+1}-v_n\sim_\infty \frac{a}{2}n^{a-1}$$ and by telescoping and repeating the same method as in $(AE)$ we have: $$v_n\sim_\infty \frac{a}{2}\int_1^n x^{a-1}dx\sim_\infty \frac{n^a}{2}$$ hence finaly we find $$\sum_{k=1}^n k^a=\frac{1}{a+1}n^{a+1}+ \frac{n^a}{2}+O\left(n^{a-1}\right)$$ and we conclude.

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From this answer you have an asymptotics $$ \sum_{k=1}^n k^a = \frac{n^{a+1}}{a+1} + \frac{n^a}{2} + \frac{a n^{a-1}}{12} + O(n^{a-3}) $$ Use it to prove that $u_n(a)$ converges.

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Norbert
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