It is easy to show $\quad\sum_{i=1}^{\infty} \dfrac{1}{n^i}=\dfrac{1}{n-1}\quad$ e.g. $$\quad\sum_{i=1}^{\infty} \dfrac{1}{2^i}=\dfrac{1}{1}\quad$ $\quad\sum_{i=1}^{\infty} \dfrac{1}{3^i}=\dfrac{1}{2}\quad$ $\quad\sum_{i=1}^{\infty} \dfrac{1}{4^i}=\dfrac{1}{3}\quad$$
and so on but what about summing to a finite power $\,n?$
$$\sum_{i=1}^{k} \dfrac{1}{n^i}=???$$
I have developed a method of deriving formulas for the sums of integer powers of any degree but this is a case of proper fractions and it is the powers that are the increasing integers in this sum. I just don't know how to approach this. Any suggestions.
$\textbf{Update:}$
Thanks to @Feng, I have been able to show that \begin{align*} \sum_{i=1}^{k} q^i&= \dfrac{q(1-q^k)}{1-q} , q=\dfrac{1}{n} \implies \dfrac{\frac{1}{n}(1-\frac{1}{n^k})}{1-\frac{1}{n}} =\frac{n^k-1}{n^k(n - 1)}, \,n\ge2 \end{align*} and Wolfram confirms that this formula works. Thank you.
I have been able to derive the finite sum of fractional powers formula (above) directly, using 3 terms to infer the workings of any finite number of terms:
\begin{align*} \sum_{i=1}^{3} \dfrac{1}{x^i} &=\dfrac{1}{x^1}+\dfrac{1}{x^2}+\dfrac{1}{x^3}\\ \\ &=\dfrac{x^2}{x^3}+\dfrac{x}{x^3}+\dfrac{1}{x^3}\\\\ &=\dfrac{x^2+x+1}{x^3}\\ \\ &=\dfrac{(x^2+x+1)}{x^3}\cdot\dfrac{(x-1)}{(x-1)}\\ \\ &=\dfrac{x^3-1}{x^3(x-1)}\\ \\ \implies \sum_{i=1}^{k} \dfrac{1}{x^i} &=\dfrac{x^k-1}{x^k(x-1)} \end{align*}