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I would like to study the convergence of the series: $$ \sum u_{n}$$ where $$ u_{n}=a^{s_{n}}$$

$$ s_{n}=\sum_{k=1}^n \frac{1}{k^b}$$ $$ a,b<1$$

We have:

$$ u_{n}=\exp((\frac{n^{1-b}}{1-b}+o(n^{1-b}))\ln(a))=\exp(\frac{\ln(a)n^{1-b}}{1-b}+o(n^{1-b}))$$

However $$\exp(o(n^{1-b}))$$ must be specified. So how can I determine: $$ o(n^{1-b})=s_{n}-\frac{n^{1-b}}{1-b}=\sum_{k=1}^n \frac{1}{k^b}-\frac{n^{1-b}}{1-b}$$ ?

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up vote 1 down vote accepted

The $o(n^{1-b})$ term would be enough to conclude, but anyway, $t\mapsto 1/t^b$ is decreasing on $t\gt0$ and $b\lt1$ hence $$ s_n\leqslant\sum_{k=1}^n\int_{k-1}^k\frac{\mathrm dt}{t^b}=\int_0^n\frac{\mathrm dt}{t^b}=\frac{n^{1-b}}{1-b}. $$ Likewise, $$ s_n\geqslant\sum_{k=1}^n\int_{k}^{k+1}\frac{\mathrm dt}{t^b}=\int_1^{n+1}\frac{\mathrm dt}{t^b}=\frac{(n+1)^{1-b}-1}{1-b}\geqslant\frac{n^{1-b}-1}{1-b}. $$ Hence, if $a\lt1$, $$ u_n=\mathrm e^{s_n\log(a)}\leqslant\exp\left(\log(a)\frac{n^{1-b}-1}{1-b}\right)=C\exp(-cn^{1-b}), $$ for some finite positive $c$ and $C$ which I will let you write down.

Edit One sees that $s_n=\dfrac{n^{1-b}}{1-b}+O(1)$ hence $u_n=\displaystyle\exp\left(\log(a)\frac{n^{1-b}}{1-b}+O(1)\right)$. Some more work yields $s_n=\dfrac{n^{1-b}}{1-b}-r_n$, where $r_n\to r$ and $r$ is finite, positive, and defined by $$ r=\int_{0}^{+\infty}\left(\frac1{t^b}-\frac1{\lceil t\rceil^b}\right)\mathrm dt. $$ Thus, $$ u_n=\displaystyle\exp\left(\log(a)\frac{n^{1-b}}{1-b}-\log(a)r-v_n\right), $$ with $(v_n)$ decreasing, $v_n\gt0$ and $v_n\to0$.

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So $$o(n^{1-b})<0$$ and $$u_{n}<\exp(\frac{\ln(a)n^{1-b}}{1-b})$$ the series converges (?) – Chon May 14 '12 at 18:14
So it is sufficient to show that $$o(n^{1-b})<0$$ ? – Chon May 14 '12 at 19:40
This (absurd) assertion is nowhere in my answer. Please check the definition of little-o. – Did May 14 '12 at 19:43
$$ o(n^{1-b})=s_{n}-\frac{n^{1-b}}{1-b}$$ And $$ s_{n} \leq \frac{n^{1-b}}{1-b}$$ So $$ o(n^{1-b}) \leq 0$$ So $$\exp(\frac{\ln(a)n^{1-b}}{1-b}+o(n^{1-b})) \leq \exp(\frac{\ln(a)n^{1-b}}{1-b})$$ Does this mean something? (although I know $o(n^{1-b}) \leq 0$ seems to be absurd!) – Chon May 14 '12 at 20:26
No reliable proof should have a step asserting that $o(n^{1-b})\leqslant0$. Once again: the best you could do in your situation is to check thoroughly the definition of little-o. Yes, this is an extremely convenient shorthand... but you use it in a questionable way. You might want to try to avoid it altogether, to see if, once expanded, your proof survives. – Did May 15 '12 at 8:48

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