Here's one way to figure out the first one. Let's consider where we get in relation to the number $2$ at each step in the sum.
The first number is $1$. That gets us halfway to $2$. The next number is $\frac 12$. This gets of half of the distance that's left. The next number is $\frac 14$. That gets of half the distance that's left again. We can see that this pattern continues (plot these points on a number line if you still don't see) so this sum will never reach $2$ after any number of finite iterations. Thus $k\lt 2$.
Here's one way to figure out the second one
$$\color{red}{n} = (1) + \left(\frac12 + \frac13\right) + \left(\frac14 + \frac15 + \frac16 + \frac17\right) \color{red}{\lt} (1) + \left(\frac12 + \frac12\right) + \left(\frac14 + \frac14 + \frac14 + \frac14\right) = \color{red}{3}$$