Probability that sum of integer reciprocals is larger than a fixed number. Suppose $n$ numbers are drawn independently from the list of $m$ integers $\{1,2,3,\ldots ,m\}$ uniformly at random. Denote these $n$ picks as $x_1,x_2,\ldots x_n$. Note that $n\geq m$ is possible. Fix a positive integer $C$. I am trying to determine the probability that $$\sum_{i = 1}^{n} \frac{1}{x_n}\geq C.$$
However I am not really sure where to start as I have not done much work with probability before. 
Is there some way to get such a probability?

This is actually more of a consideration than an answer, 
but wishfully it may be of some help.
We have $n$ discrete uniform i.i.d. random variables $X_k$, ranging from $1$ to $m$, 
and we want to find the distribution of the sum of their inverse.
To the scope of finding an approximation for high values of $m$ and $n$,  we should go through 
the Characteristic Function of each variable $1/X_k$, exploiting the fact that the CF of
the sum will be the $n$-th power of the single CF. And after getting the global CF we can invert it to get the sought pdf.
The single CF is given by
$$
\eqalign{
  & \varphi _{1/X} (t) = E\left[ {e^{\,i\,t/X} } \right] = {1 \over m}\sum\limits_{k = 1}^m {e^{\,i\,t/k} }  =   \cr 
  &  = e^{\,i\,t} {1 \over m}\sum\limits_{k = 1}^m {e^{\, - i\,t\left( {k - 1} \right)/k} }  \cr} 
$$
and the matter comes to find a suitable approximation for the sum, or rather for its logarithm.
It might help
to approximate each variable with
a continuous uniform one ranging, from $1/2$ to $m+1/2$ and with probability density 
of $1/m$.
Geometrically that means to approximate a discrete hystogram of $m$ bars of height $1/m$
with $m$ vertical bands (rectangles) of width $1$ and height $1/m$, centered 
around each integral point.
Then the Characteristic Function of each continuous variable $1/X_k$ would be
$$
\eqalign{
  & \varphi _{1/X} (t) = E\left[ {e^{\,i\,t/X} } \right] = {1 \over m}\int_{\;x = 1}^{\,m} {e^{\,i\,t/x} dx}  =   \cr 
  &  = {1 \over m}\left( {\int_{\;x = 1}^{\,m} {\cos \left( {{t \over x}} \right)dx}  + i\int_{\;x = 1}^{\,m} {\sin \left( {{t \over x}} \right)dx} } \right) \cr} 
$$
