I will try to get control of the most standard interpretation of our question by using (at first) very informal language.  Let us call someone *unhappy* if one or more people share his/her "birthday."  We want to find the "expected number" of unhappy people.

Define the random variable $X$ by saying that $X$ is the number of unhappy people.
We want to find $\text{E}(X)$.  Let $p_i$ be the probability that $X=i$.  Then 
$$\text{E}(X)=\sum_{i=0}^{n} i\,p_i$$ 
That is roughly the approach that you took.  That approach is correct, and a very reasonable thing to try.  Unfortunately, in this problem, finding the $p_i$ is very difficult.  One could, as you did, decide that for a good approximation, only the first few $p_i$ really matter.  That is sometimes true, but depends quite a bit on the values $N$ of "days in the year" and the number $n$ of people.

Fortunately, in this problem, and many others like it, there is an alternative *very* effective approach.  It involves a bit of theory, but the payoff is considerable.

Line the people up in a row.  Define the random variables $U_1,U_2,U_3,\dots,U_n$ by saying that $U_k=1$ if the $k$-th person is unhappy, and $U_k=0$ if the $k$-th person is not unhappy.  The crucial observation is that
$$X=U_1+U_2+U_3+\cdots + U_n$$  

One way to interpret this is that you, the observer, go down the line of people, making a tick mark on your tally sheet if the person is unhappy, and making no mark if the person is not unhappy.  The number of tick marks is $X$, the number of unhappy people.  It is also the sum of the $U_k$.

We next use the following very important theorem: **The expectation of a sum is the sum of the expectations**.  This theorem holds "always."  The random variables you are summing *need not be independent*.  In our situation, the $U_k$ are not independent, but, for expectation of a sum, that does not matter. So we have
$$\text{E}(X)=\text{E}(U_1) + \text{E}(U_2)+ \text{E}(U_3)+\cdots +\text{E}(U_n)$$

Finally, note that the probability that $U_k=1$ is, as carefully explained by @Henry, equal to $p$, where
$$p=1-(1-1/N)^{n-1}$$
It follows that $\text{E}(U_k)=p$ for any $k$, and therefore $\text{E}(X)=np$.