用极限定义无穷级数

Previously, we began to talk about infinite series, and how they behave. Here, we'll continue that discussion with an emphasis on the limit definition of an infinite series.

Let's start with a sequence $\{a_n\}_{n=1}^{\infty}$. Define $$s_n=\sum_{i=1}^n a_i.$$ This gives us a new sequence $\{s_n\}_{n=1}^{\infty}$, where

$$s_1=a_1,$$ $$s_2=a_1+a_2,$$ $$s_3=a_1+a_2+a_3,$$

and so on. (These $s_i$'s are called partial sums.) With this notation, we define the infinite sum as $$\sum_{i=1}^{\infty} a_i=\lim_{n\to\infty} s_n.$$ Furthermore, we say that the series is convergent if the limit on the right side exists, and otherwise we call it divergent. In the convergent case, if $\lim_{n\to\infty} s_n=S$, then we also say that $\sum_{i=1}^{\infty} a_i=S$ as well.

Using this definition, we can figure out whether certain infinite series converge or diverge. For example, let's see if the infinite sum $$\sum_{i=1}^{\infty} \frac{1}{2^n}$$ converges. To do so, we set the original sequence as $a_i=\frac{1}{2^i}$, which turns our infinite series into $$\sum_{i=1}^{\infty} \frac{1}{2^n}=\sum_{i=1}^{\infty} a_i=\lim_{n\to\infty} s_n,$$ where we define $$s_n=\sum_{i=1}^n \frac{1}{2^i}=\frac{1}{2^1}+\frac{1}{2^2}+\cdots+\frac{1}{2^n}.$$ We see that this is a finite geometric series with ratio $r=\frac{1}{2}$. So, we can apply the appropriate formula to find that $$s_n=a_1\cdot\frac{1-r^n}{1-r}=\frac{1}{2}\cdot\frac{1-\left(\frac{1}{2}\right)^n}{1-\frac{1}{2}}=1-\frac{1}{2^n}.$$ Therefore, since $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} \left(1-\frac{1}{2^n}\right)=1,$$ by the limit definition, the infinite series converges and equals $1$. In other words, $$\sum_{i=1}^{\infty} \frac{1}{2^i}=1.$$

Let's look at another example together, and see whether the series $$\sum_{i=1}^{\infty} \frac{1}{i(i+1)}$$ converges. Again, as before we set $a_i=\frac{1}{i(i+1)}$ and try to find a closed expression for $$s_n=\sum_{i=1}^n \frac{1}{i(i+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{n(n+1)}.$$

We observe that we can write $$\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}.$$ So, it follows that $$s_n=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}.$$ And since $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} \left(1-\frac{1}{n+1}\right)=1,$$ we conclude that the original infinite sum converges and equals $1$.

Now try the next problem on your own!

For this problem, our initial sequence is given by the equation $a_i=i$. So, we define the partial sums as $$s_n=\sum_{i=1}^n i=1+2+\cdots+n.$$

Suppose we take the sum $$s_n=1+2+\cdots+(n-1)+n$$ and rearrange it into $$s_n=n+(n-1)+\cdots+2+1.$$ Now, let's add the two equations together! The trick here is that we don't combine all terms at the same time, but we add each pair separately: $$2s_n=[1+n]+[2+(n-1)]+\cdots+[(n-1)+2]+[n+1].$$ We notice that each term in the brackets is equal to $n+1$. So, since there are $n$ total pairs, $$2s_n=(n+1)+(n+1)+\cdots+(n+1)=n(n+1).$$ So, we conclude that $$s_n=\frac{n(n+1)}{2}.$$ Now that we have a closed expression for the $n$-th partial sum, we need to consider the limit $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} \frac{n(n+1)}{2}.$$ By the limit definition, the original infinite series converges if and only if this limit exists! However, in this case we know that $$\lim_{n\to\infty} \frac{n(n+1)}{2}=\infty.$$ Therefore, we conclude that the infinite series $\sum_{i=1}^{\infty} i$ diverges.

We end with one note: while we were able to determine whether the thre series above converge or diverge, we had to use additional formulas. For the first series, we had to know the formula for the sum of a finite geometric series; for the second, we needed to know how to evaluate telescoping sums; and for the third, we had to derive and apply a relatively random formula for the sum of the first $n$ natural numbers. As we will continue to see, using the limit definition often proves to be a much longer or more difficult method of testing convergence, which is why we will also learn about more convenient tests for convergence in the following sections.

Using the Limit Definition to Test for Series Convergence

How do you determine whether an infinite series converges or diverges? In the simplest cases, you can stick to the basics and just directly use the limit definition of infinite series. Sometimes, like for infinite geometric series, you might be able to evaluate the series exactly (if it converges), not just prove convergence or divergence!