Sequence of partial sums of e in Q is a Cauchy sequence.

Question

Verify that $X_n= \{ \sum_{i=0}^n$ $\frac{1}{i!}$} is a Cauchy sequence in $Q$ with the Euclidean metric.

I can't figure out how to find an $N$ that makes this work. I figure that $d(x_n,x_m) < \frac{n-m}{m!}$ if $m \neq n$ and $n > m$ , but beyond that I'm lost.

Answer 1

HINT: You don’t need to find $N$ explicitly. Fix $\epsilon>0$. Since $\sum_{k\ge 0}\frac1{k!}=e$, you know that there is an $N\in\Bbb N$ such that

$$\left|e-\sum_{k=0}^N\frac1{k!}\right|=e-\sum_{k=0}^N\frac1{k!}=\sum_{k\ge N}\frac1{k!}<\frac{\epsilon}2\;.$$

The terms of the series are positive, so for each $n\ge N$ we have

$$|e-X_n|=e-X_n=\sum_{k\ge m}\frac1{k!}<\frac{\epsilon}2\;.$$

If $n,m\ge N$, then $|X_n-X_m|=|(e-X_m)-(e-X_n)|$. Now use the triangle inequality.

Answer 2

Denote $ a_k = 1/k! $ and note that we have $ \lim_{k\to \infty} a_{k+1}/a_k = 0 $. The ratio test then implies the result; but let's write it out more explicitly. Fix $\epsilon > 0 $. By the above limit, it follows that there is $ N_1 $ such that for all $ k > N_1 $ we have $ a_{k+1} < a_k/2 $, from which it follows that $ a_{k+r} < a_k/2^r $. We then have, for $ n > m > N_1 $:

$$ d(X_n, X_m) \leq \sum_{k=m}^{n-1} d(X_{k+1}, X_k) = \sum_{k=m}^{n-1} a_{k+1} = \sum_{k=0}^{n-m} a_{m+k} < a_m \sum_{k=0}^{n-m} 2^{-k} < 2a_m $$

But since $ \lim_{m\to \infty} a_m = 0 $, there is $ N_2 $ such that for all $ m > N_2 $ we have $ a_m < \epsilon/2 $. Picking $ N = \max\{N_1, N_2\} $ and imposing $ n, m > N $ then yields that $ d(X_n, X_m) < \epsilon $ and we are done.