What is the difference betweeen Riemann sums and series?

Question

If we check a function $f:[a,b] \to \mathbb{R}$ for Riemann-integrability we usally take a closer look at the upper and lower Riemann-sums and try to figure out if they converge (in terms of refinements).

I will use the following example to explain my problem:

Let be $f:[a,b] \to \mathbb{R}$, where $f(x)=x$. If we show that $f$ is Riemann-integrable I would proceed as follows:

Let be $P_n=(a, t_1, t_2, ...,t_{n-1}, b)$ a partition of $[a, b]$ where $t_i=a+\frac{b-a}{n}i$ and $i \in \{0, 1, ..., n\}$. We define $m_i:= inf\{f(x),~x \in [t_i, t_{i+1}]\}$ and $M_i:= sup\{f(x),~x \in [t_i, t_{i+1}]\}$. Hence, for this particular partition the Riemann-sums are:

$L(f(x),P_n)= \sum^n_i m_i (t_{i+1}-t_i)$ and $U(f(x),P_n)= \sum^n_i M_i (t_{i+1}-t_i)$.

We want to show that there exists a refinement $P^*$ of $P_n$ such that $U(f(x),P^*) - L(f(x),P^*)=0$. Now we take the difference $U(f(x),P_n)-L(f(x),P_n)$ and after some algebraic manipulations we get: $(b-a)\frac{b-a}{n}$. If $n \to \infty$, which is something like refining the partition, we see that the difference converges to $0$. This means that there exists such a $P^*$. Further, we know from lecture that for all other partitions $P'$ (in particular ones that are not constructed in the way I did) which are finer than $P^*$ we also have $U(f(x),P') - L(f(x),P')=0$.

---

At this step I am a little bit confused. The Riemann-sums look like function series. So I am wondering if it is legit to argue with convergence criteria of function series when it comes to convergence of Riemann-sums?

How can I explain the connection, if there is any, between Riemann-sums and "ordinary" function series?

Please let me know if this question is too vague, otherwise I would appreciate any comments or explanations to clarify my thoughts on that.

Answer 1

No, convergence criteria of series does not apply here. For each partition $P$, $U(f,P)$ and $L(f,P)$ are finite sums. And, if you have partitions $P_1,P_2,P_3,\ldots$, each of which is a refinement of the previous one, the sequences $\bigl(U(f,P_n)\bigr)_{n\in\mathbb N}$ and $\bigl(L(f,P_n)\bigr)_{n\in\mathbb N}$ are not partial sums of a fixed series.

Answer 2

You're seeing a lot of finite sums, and thinking "there must be a series". But, the partial sums that converge to an infinite series are just different beasts from the Riemann sums that converge a Riemann integral.

Let's consider a series, $\sum_{i=1}^\infty a_i$.

It's first partial sum is $$S_1 = \sum_{i=1}^1 a_i = a_1 $$ It's second partial sum is $$S_2 = \sum_{i=1}^2 a_i = a_1 + a_2 = S_1 + a_2 $$ It's third partial sum is $$S_3 = \sum_{i=1}^3 a_i = a_1 + a_2 + a_3+ = S_2 + a_3 $$ It's fourth, and fifth, and sixth partial sums are $$S_4 = S_3 + a_4, \qquad S_5 = S_4 + a_5, \qquad S_6 = S_5 + a_6 $$ In general, each subsequent partial sum $S_{n+1}$ is obtained from the previous $S_n$ by adding just a single term of the series, namely $a_{n+1}$. And, convergence of the series, by definition, means convergence of the sequence of partial sums, and each partial sum is obtained.

Welp, this is not what's going on with a Riemann sum. Each subsequent Riemann sum is a brand new sum formed from brand new terms, not related to the terms of the previous Riemann sum.