I have the next question, if we take $t \in \mathbb{R}$ (maybe $t > 1$) we can estimate the tail of the serie $\sum_{n} \frac{1}{n^t}?$ for the $n \geq K$, $K$ fix positive integer.
tail estimate for ''real zeta function''
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2See the proof [here](http://en.wikipedia.org/wiki/Integral_test_for_convergence), or the "Integral Test" section [here](http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx). – 2012-08-02
1 Answers
Yes, we can estimate the tail of the series. Here is how :
It is proved that if $\sum{a_n}$ is convergent and $f(n)=a_n$ where $f$ is a continuous, positive, decreasing function for $x\geq K$, then the tail $T_K=\sum_{n={K+1}}^\infty a_n$ of the series satisfy $ \int_{K+1}^\infty f(x)dx\leq T_k \leq\int_K^\infty f(x)dx $
We can use this theorem for $\sum_{n}\frac{1}{n^t}$ with $t>1$ since the series is then convergent and $f(x)=\frac{1}{x^t}$ satisfy the requirements. You can prove its convergence for example with the Integral Test as proposed by David Mitra.
Hence we compute $ \int_{K+1}^\infty\frac{1}{x^t}dx=\frac{1}{(t-1)(K+1)^{t-1}},\quad \text{and}\quad\int_K^\infty\frac{1}{x^t}dx=\frac{1}{(t-1)K^{t-1}} $ to get $ \frac{1}{(t-1)(K+1)^{t-1}} \leq T_k \leq \frac{1}{(t-1)K^{t-1}} $ so that we can approximate the tail by taking the middle value : $ T_k\simeq \frac{1}{2}\left(\frac{1}{(t-1)(K+1)^{t-1}}+\frac{1}{(t-1)K^{t-1}} \right)=\frac{(K+1)^{1-t}+K^{1-t}}{2(t-1)} $