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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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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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    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.2012-05-15