Basic doubt about Infinite Series and the $S_{n}$ term My question is: why, in general we cannot write down an formula for the $n-$th term, $S_{n}$, of the sequence of partial sums?
I will explain better in the following but the question is basically that one above.
Suppose then you have an infinite sequence in your pocket, $\{a_{1},a_{2},a_{3},...\}$, or,
$$\{a_{1},a_{2},a_{3},...\} \equiv \{a_{n}\}_{n=0}^{\infty} \tag{1}$$
$(1)$ then is a fundamental object because then you can "sum up" all the terms of this particular sequence, just like: $a_{0}+a_{1}+a_{2}+\cdot \cdot \cdot$ to define another object. Well, doing that procedure you construct that object, called infinite series of the infinite sequence $\{a_{n}\}_{n=0}^{\infty}$ 
$$a_{0}+a_{1}+a_{2}+\cdot \cdot \cdot \equiv \sum^{\infty}_{n=0}a_{n} \tag{2}$$
The next procedure you might like to do is then question yourself if a infinite series have some value $s \in \mathbb{K}$ ($\mathbb{K}$ a field) indeed. The procedure to answer that question is then firstly construct another infinite sequence called the sequence of partial sums of the series:
$$\{S_{0},S_{1},S_{2},S_{3},...,S_{k},...\} \equiv \{S_{n}\}_{n=0}^{\infty} \tag{3}  $$
Which is:
$$\begin{cases} S_{0} = \sum^{0}_{n=0}a_{n} = a_{0}\\S_{1} = \sum^{1}_{n=0}a_{n} = a_{0} + a_{1} \\ S_{1} = \sum^{2}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} \\ S_{3} = \sum^{3}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} + a_{3} \\\vdots\\ S_{k} = \sum^{k}_{n=0}a_{n} = a_{0} + a_{1} + a_{2} + a_{3}+\cdot \cdot \cdot+a_{k}\\ \vdots  \end{cases} $$
and then calculate the limit of this sequence $(3)$, like:
$$ \lim_{n\to \infty} \sum^{n}_{j=0}a_{j} \equiv \lim_{n\to \infty} S_{n} \tag{4} $$
Now, if the limit $(4)$ has a value $s = L$ then the can say that the Sum of the Series is that limit:
$$ \sum^{\infty}_{n=0}a_{n} = s  \tag{5}$$
$$ * * * $$
Now, if we do not have a proper expression for $S_{n} = \sum^{k}_{n=0}a_{n}$, then the whole "direct limit calculus" do not work and then we need other methods for search the value (more generally the convergence) of a series (e.g. integral test). The thing is, I do not see (understand) why we cannot in general write down a formula for $S_{n}$ and some times we can. For instance, I do not see why in one hand we can write down a formula for geometric series but on the other hand we cannot for harmonic series, for me the $S_{n}$ term, of the harmonic series, to plug up in the limit is given by:
$$ S_{n} = \sum^{n}_{k=0}\frac{1}{n} = 1+ \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} \equiv \Big( 1+ \frac{1}{2} + \frac{1}{3} + ... \Big) + \frac{1}{n} =    $$
$$= C + \frac{1}{n} $$
$C$ a constant since it's a finite sum. Then,
$$\lim_{n\to \infty} C + \frac{1}{n} = C$$
Then,
$$\sum^{\infty}_{n=0}\frac{1}{n} = C$$
I know that what I wrote above isn't write, but I simply do not understand why. There's a subtle thing that I do not understand. Anyway, the question is posted above.
Thank You.   
 A: Yes, you are right: in general we cannot write down a formula for the $n$th partial sum of a sequence. As there is no simple closed expression for most primitives, such as $\int\frac1{\log(x)}\,\mathrm dx$, $\int e^{x^2}\,\mathrm dx$, and so no. And, in general, there is no simple closed expression for $\prod_{k=1}^na_k$. There is nothing peculiar about series here.
A: You had to write $C_n$ instead of $ C$.
In fact we have
$$C_n=S_{n-1} \text{ and } \; S_n=C_n+\frac 1n$$
and all we can say is
If $(C_n)$ converges then the series is convergent.
A: Actually you can express the partial sums of the Harmonic sequence
$$
\sum\limits_{k = 1}^n {{1 \over k}}  = \psi (n + 1) - \psi (1) = \gamma  + \psi (n + 1)
$$
through the Digamma Function.
For the general question Josè answered already.
A: So, the answer depends on what you are willing to consider as a solution.  If I understand your question, you are asking why, if it is rational to think of a particular series, then it is also rational to think of its partial sums, and there should be no reason we cannot deduce a general formula for the partial sum from $k = 1$ to $n$ that doesn't include counting, since it is literally a function of $n$.  That is, the result is (a) fixed, and (b) a function of $n$, $f(n)$, so why can't there be a formula for it?
The answer, in short, is based on what you consider to be a valid list of symbols in your function.  For instance, let's say that we hadn't discovered exponentiation yet.  In such a case, we would not be able to write the result of the geometric series as a formula, right?  Therefore, the functions that we are able to write a formula for is actually a function of the operators that we know about (or are willing to include).
Ultimately, any given function can be written as a formula, if you are willing to make the function itself a standard operator.  So, if I have a series $S$ with a formula for the $k$th term $g(k)$, I can define a function $Q(n)$ to represent the partial sum to $n$, such that $Q(n) = \sum\limits_{k = 1}^n g(k)$.  You might ask me how I am going to represent this as a formual.  Well, to some extent, I already have!  $Q(n)$ IS a representation of this as a formula, if I allow it into my list of allowable functions for representation.
In short, we need to recognize that the functions that we know about, use, and allow in our formulas have some amount of arbitrariness to them, many are historically contingent on the functions that humans have found interesting.  Therefore, the list of formulas that can be modeled using these sub-formulas will be likewise limited.  You can always expand this list if you want to, and if you add in the right operators to your list you will be able to represent the function you are seeking, since you could always add the function itself as one of your allowed functions.
