It has been a really long time since I had to solve a problem like this and I was wondering if I could get some assistance with it.
$$x= \sum_{i=1}^{t} \frac{1}{i^c} \hspace{.25cm} \text{where } c \geq 0 $$
I am trying to solve for t for a given x. If I recall correctly, one of the first steps is to write out the series like:
$$x = \frac{1}{1^c} + \frac{1}{2^c} +\frac{1}{3^c} + .... + \frac{1}{t^c}$$
but I haven't a clue how to continue after that.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} x & \equiv \bbox[5px,#ffd]{\left.\sum_{i=1}^{t} \frac{1}{i^c} \,\right\vert_{\ds{\ c \geq 0}}} = \zeta\pars{c} + {t^{1 - c} \over 1 - c} + c\int_{t}^{\infty}{\xi - \left\lfloor \xi\right\rfloor \over \xi^{\,c\ +\ 1}}\,\dd\xi \end{align} which is a Zeta function-$\ds{\zeta}$ identity.
twithout having to actually iterate and continuously check againstx. – Jedi_Maseter_Sam Oct 02 '20 at 17:55