Proof that arithmetic series diverges

Question

Let $(a_n)_{n=1}^\infty$ be a sequence where for any $i$, we have $a_{i+1}-a_i=d\in\mathbb{R}$.

It is intuitively obvious that the series $a_1+a_2+a_3+\dots$ diverges (unless all $a_i$ are identically zero). But when I try to write down a formal proof of this, I end up with a surprisingly long and inelegant proof.

So I'm wondering if anyone knows of a simple proof.

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The definition of convergence I'm using:

We say that $a_1+a_2+a_3+\dots$ converges to a real number $L$ if for all $\epsilon>0$, there exists $N$ such that for all $k\geq N$, $|\sum_{i=1}^ka_i - L|<\epsilon$. Otherwise, it diverges.

Ideally I would like to prove the above by directly using this definition (and without resort to other tests like the Term Test or the Cauchy criterion).

Answer 1

If $\sum_{i=1}^\infty a_i$ converges then necessarily $\lim_{i\to \infty} a_i = 0$ (see term test).

It follows that $d = a_{i+1}-a_i \to 0 $ and therefore $d=0$.

So all $a_i$ are equal, and since their limit is zero, all $a_i$ are equal to zero.

For a direct proof (although that hides the essential idea), write $$ \begin{align} d &= a_{k+1} -a_k = \sum_{i=1}^{k+1} a_i - 2 \sum_{i=1}^{k} a_i + \sum_{i=1}^{k-1} a_i \\ &= \left(\sum_{i=1}^{k+1} a_i - L \right)- 2 \left(\sum_{i=1}^{k} a_i - L \right)+ \left(\sum_{i=1}^{k-1} a_i- L \right) \end{align} $$ and note that – given $\epsilon > 0$ – the absolute value of the right-hand side is less than $4 \epsilon$ for sufficiently large $k > N$, so that $d$ must be zero.

Answer 2

The $n$-th term of the sequence is given by $a_n=a_1+(n-1)d$ and so if $d\geq0$ $$\sum_{n=1}^ma_n=\sum_{n=1}^m(a_1+(n-1)d)\geq na_1,$$ which shows that the series diverges if $a_1>0$. A similar argument works for $d<0$ and $a_1<0$.

Note that the series does converge if $a_1=d=0$.

Answer 3

The partial sums are a non-constant polynomial in $k$ and polynomials are known to be unbounded.

Answer 4

If you want a proof directly from the definition, the definition of a limit is that there is some $L$ such that given any $\epsilon$, there is some $N$ such that every term after $a_N$ is within $\epsilon$ of the limit. So the negation of that would be that given any supposed limit $L$, an $\epsilon$ can be found such that for any $N$, there is a term after $a_N$ further than $\epsilon$ from $L$.

So suppose someone choose an $L$. We can easily find a term that is greater than $L$. For instance, take $k = \lceil \frac {L-a_0}d \rceil +1 $ (this assumes that $d>0$; if $d<0$, this can be easily modified). Simple algebra show that $a_k \ge L+1$. Since $a_n$ is increasing (again, assuming $d>0$), we also have $a_n \ge L+1$ for any $n>k$. We can rearrange that to $a_n -L \ge 1$. Choosing any $\epsilon$ less than $1$ will result in $|a_n-L|>\epsilon$.

Now we just have to show that given any $N$, there is some $n>N$ such that $|a_n-L|>\epsilon$. To do so, we just have to take some $n>\max(N,k)$.

Answer 5

It's necessary for the terms to vanish at infinity if the sum is to converge to a finite value. No arithmetic progression has this property. Indeed, for all such progressions, no matter how small the difference, the terms grow without bound. Thus, their sum also cannot be finite.