Answer
$\color{blue}{\frac{2}{n+1}}$
Work Step by Step
Let $f_X(x)$ be the pdf of $X$. Then,
$\quad f_X(i)= P(X=i) = ki,\ i=1,2,3,\ldots, n,$
so that
\begin{align*}
\sum\limits_{\Omega_X} f_X(x) &= 1 \\
\sum_{i=1}^n f_X(i) &= 1 \\
\sum_{i=1}^n ki &=1 \\
k\sum_{i=1}^n i = 1 \\
k\cdot \frac{n(n+1)}{2} &= 1 \\
k &= \frac{2}{n(n+1)}.
\end{align*}
Thus,
$\quad f_X(i) = \dfrac{2}{n(n+1)}i,\ i=1,2,3,\ldots, n.$
Consequently,
\begin{align*}
E\left(\frac{1}{X}\right) &= \sum\limits_{\Omega_X} \frac{1}{x} \cdot f_X(x) \\
&= \sum_{i=1}^n \frac{1}{i}\cdot f_X(i) \\
&= \sum_{i=1}^n \frac{1}{\not{i}}\cdot \frac{2}{n(n+1)}\not{i} \\
&= \sum_{i=1}^n \frac{2}{n(n+1)} \\
&= \frac{2}{n(n+1)} \sum_{i=1}^n 1 \\
&= \frac{2}{\not{n}(n+1)} \cdot \not{n} \\
\color{blue}{E\left(\frac{1}{X}\right)}\ &\color{blue}{= \frac{2}{n+1}.}
\end{align*}
