Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathit{e} - s_n$.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer

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:

$$\sum_{i=n+1}^\infty a_i$$

share|cite|improve this answer

The symbol $\sum_{k=0}^\infty a_k$ is used for two diferent things:

  1. The sequence of partial sums $\{\sum_{k=0}^na_k\}$.
  2. 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:

  1. The sequence $\{\sum_{k=0}^n\frac{1}{k!}\}$ converges as $n\to\infty$.
  2. The limit of the above sequence is the number $e$.

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

  1. The sequence $\{\sum_{k=n+1}^m\frac{1}{k!}\}$ converges as $m\to\infty$.
  2. The limit of the above sequence is $e-s_n$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.