written one way, the limit doesn't exist. written another, the limit equals 0?

Question

Hi I'd greatly appreciate some help/clarification with this problem, I'm also open to suggestions about how to be more articulate when writing problems out in general.

On my way to saying that the series $\sum_{n=1}^{\infty}(-1)^{n+1}cos(\pi/n)$ diverges,

I figured I would show that $lim_{n\rightarrow\infty}(-1)^{n+1}cos(\pi/n)\neq0$

I tried re-writing the expression $(-1)^{n+1}cos(\pi/n)$ as $cos(\pi/(2n-1))-cos(\pi/2n)$,

because I figured I could just show that $lim_{n\rightarrow\infty}cos(\pi/(2n-1))-cos(\pi/2n)$.

But that limit equals 0.

I suppose I wonder

(1) why If $\sum a_n=\sum b_n$, why can't we assume that $lim_{n\rightarrow\infty} a_n=lim_{n\rightarrow\infty} b_n $ ?

and

(2) Maybe (1) does hold, and I am confused about the first limit? I think it equals $\pm1$.

Answer 1

(this first part just added)

Here's as general as I can make it:

Let $(a_n)_{n=0}^{\infty}$ be a sequence of reals (or complex numbers) such that $L = \lim a_n$ exists and $\sum_{n=0}^{\infty} |a_n-L| $ converges. Let $(c_k)_{k=0}^{m-1}$ be real or complex numbers such that $\sum_{k=0}^{m-1} c_k = 0 $ and $\sum_{k=0}^j c_k \ne 0 $ for $0 \le j < m-1$.

Then the sum $\sum_{k=0}^{\infty} a_k c_{k\bmod m} $ can have $m$ distinct values, each gotten by first summing the first $j$ terms of the sum for $j = 1$ to $m$, and the remaining terms of the sum being grouped in sets of $m$ terms.

That is, for $j = 1$ to $m$, $S_j =\sum_{k=0}^{j-1} a_k c_{k} +\sum_{n=0}^{\infty}\sum_{k=0}^{m-1} a_{mn+j+k} c_{(k+j)\bmod m} $ exists and is a possible value for $\sum_{k=0}^{\infty} a_k c_{k\bmod m} $.

This is done by showing that, if $b_{n, j} =\sum_{k=0}^{m-1} a_{mn+j+k} c_{(k+j)\bmod m} $, then the limit conditions on the $a_n$ and the fact that the sum of the $c_m$ is zero implies that $\sum_{n=0}^{\infty} b_{n,j}$ converges.

(now back to the original answer)

If $\lim a_n =L$ exists and is non-zero, $\sum_{n \ge 1} (-1)^n a_n$ can be written in two ways: $\sum_{n \ge 1}(- a_{2n-1}+a_{2n})$ and $-a_1+\sum_{n \ge 1}(- a_{2n+1}+a_{2n})$.

If those last two sums exist, there may be two possible values for $\sum_{n \ge 1} (-1)^n a_n$.

An example is $a_n = 1-\frac1{n}$.

(added) An even simpler example is the traditional $a_n = 1$, where the two sums are always zero, so the possible answers are $0$ and $1$.

I believe there is an article in a recent MAA magazine about this.

Answer 2

The condition $a_n\to 0$ is necessary for the convergence of the series $\sum_{n\ge 1}a_n$. Therefore, you series does not converge - $|(-1)^n \cos (\pi/n)|\to 1$ as $n\to \infty$.

As a consequence, if both series $\sum a_n$ and $\sum b_n$ converge, then $\lim a_n = \lim b_n=0.$