Infinite Finite Difference

Relevant wiki: Cauchy Sequences

The first step is to note that $a_n$ must be monotone decreasing. To see this, assume that we had $a_{n+1} > a_n$ for some $n$ . Then, $c_n = a_n - a_{n+1} < 0$ , and since the sequence $c_n$ is nonincreasing its limit has to be strictly smaller than 0, which contradicts the fact that

$\lim_{n\to \infty} c_n = \lim_{n\to \infty} a_n - \lim_{n\to \infty} a_{n+1} = 0 - 0 = 0$

Denote $d_n = n c_n$ . The sequence $a_n$ is convergent, so in particular it is a Cauchy sequence. This means that for any $\epsilon > 0$ , there is $N$ such that for all $n > N$ , we have that $|a_{2n} - a_n| < \epsilon$ . Using monotonicity and positivity of $a_n$ , this gives

$\epsilon > |a_{2n} - a_n| = a_n - a_{2n} = \sum_{k=n}^{2n-1} a_k - a_{k+1} = \sum_{k=n}^{2n-1} c_k \geq n c_{2n-1}$

Likewise, for the same $N$ we have that $|a_{2n+1} - a_n| < \epsilon$ , and an entirely similar calculation shows that $(n+1) c_{2n} < \epsilon$ . Now, from these it follows that

$\lim_{n\to \infty} d_{2n} = 2\lim_{n\to \infty} n c_{2n} = 0$ $\lim_{n\to \infty} d_{2n+1} = 2\lim_{n\to \infty} n c_{2n+1} + \lim_{n\to \infty} c_{2n+1} = 0 + 0 = 0$

Thus, the desired result $\lim_{n\to \infty} d_n = 0$ follows, since any term of the sequence $d_n$ belongs to one of the above subsequences.