Subtracting the partial sums from the limit of partial sums (where the limit exists) I have always been puzzled by the phenomenon:
We know that $\mathit{e} = \sum_{k=0}^\infty \frac{1}{k!}$, and let $s_n= \sum_{k=0}^n \frac{1}{k!}$ the partial sum of $\mathit{e}$. We also know that this series converges.
I saw in some proof you can actually take $\mathit{e} - s_n= \sum_{k=n+1}^\infty \frac{1}{k!}$. I always thought that the $\sum_{k=0}^\infty$ is just a symbol that signifies that you let the partial sum tends to infinity and that's it so I do not really know how to justify the operation of $\mathit{e} - s_n$.
 A: First, "we also know that this series converges" isn't quite right, because the equation $\mathrm e=\sum_{k=0}^\infty\frac1{k!}$ already implies that the series converges.
You're right to a certain extent that $\sum_{k=0}^\infty$ is "just a symbol", but it's not just a symbol that signifies that you do something, it's also a symbol that denotes the result of doing that: It stands for the limit of the series. Thus, written out more explicitly, your equation is
$$
\lim_{m\to\infty}\sum_{k=0}^m\frac1{k!}-\sum_{k=0}^n\frac1{k!}=\lim_{m\to\infty}\sum_{k=n+1}^m\frac1{k!}\;.
$$
This you can prove directly using the fact that the limit of the difference of two convergent sequences is the difference of the limits.
A: In general, if $\sum_{i=0}^\infty a_i$ and $\sum_{i=0}^\infty b_i$ both converge, then the series $\sum_{i=0}^\infty (a_i-b_i)$ converges to the difference.
In particular, if $b_i=a_i$ for $i=0,...,n$ and $b_i=0$ for $i>n$, you get that $\sum_{i=0}^\infty b_i = \sum_{i=0}^n a_i$.  So the difference:
$$\sum_{i=0}^\infty a_i - \sum_{i=0}^n a_i = \sum_{i=0}^\infty (a_i-b_i)$$
But $a_i-b_i=0$ for $i\leq n$ and $a_i-b_i=a_i$ for $i>n$. So this sum is:
$$\sum_{i=n+1}^\infty a_i$$
A: The symbol $\sum_{k=0}^\infty a_k$ is used for two diferent things:


*

*The sequence of partial sums $\{\sum_{k=0}^na_k\}$.

*The limit as $n\to\infty$ of the above sequence if it exists. In that case, the series is said to converge, and $\sum_{k=0}^\infty a_k$ is caled the sum of the series.


This introduces some ambiguity, bit in most cases it is clear what is meant.
When you write $e=\sum_{k=0}^\infty\frac{1}{k!}$ you are conveying two facts:


*

*The sequence $\{\sum_{k=0}^n\frac{1}{k!}\}$ converges as $n\to\infty$.

*The limit of the above sequence is the number $e$.


Similarly, $e-s_n=\sum_{k=n+1}^\infty\frac{1}{k!}$ means


*

*The sequence $\{\sum_{k=n+1}^m\frac{1}{k!}\}$ converges as $m\to\infty$.

*The limit of the above sequence is $e-s_n$.

