Show that $X_{n}=\sum_{k=0}^{n} \frac{1}{k!}$ and $X_{n}=\sum_{k=0}^{n} \rho^{k}$ are Cauchy sequences.

Question

Show that $X_{n}=\sum_{k=0}^{n} \frac{1}{k!}$ and $X_{n}=\sum_{k=0}^{n} \rho^{k}$ are Cauchy sequences.

I know that for each sequence I have to show the following:

"$\forall \epsilon>0 \text{ }\exists \text{ } N_{\epsilon}=N(\epsilon) \in \mathbb{N}$ s.t. $\forall n \geq N_{\epsilon}$ , $ \left | X_{n}-X_{m} \right | < \epsilon $"

1) For the first I have done the following:

$\left | X_{n}-X_{m} \right |=\left | \sum_{k=0}^{n} \frac{1}{k!}-\sum_{k=0}^{m} \frac{1}{k!} \right |= \left | \sum_{k=n+1}^{m} \frac{1}{k!}\right | $

After this I want to use the fact that $\frac{1}{k!}\leq \frac{1}{2k}$ for $k\geq 3$, suppose that $m\geq n$, and get:

$= \left | \sum_{k=n+1}^{m} \frac{1}{k!}\right | \leq \left | \sum_{k=n+1}^{m} \frac{1}{2k}\right |=\left | \frac{1}{2(n+1)}+\frac{1}{2(n+2)}+\cdots +\frac{1}{2(m)}\right |$

then I have to bound this with a value that depends on n, on which I will impose that it is less or equal to $\epsilon$ and deduct the value of $N_{\epsilon}$ from there.

my guess so far is $\frac{m-n}{n}<\frac{m}{n} <\epsilon$

but this wont work because it is too big and I it should be "sufficiently" small (smaller than $\epsilon$).

2) For the second sequence I have done the following:

Since $0<\rho<1 \Rightarrow \sum_{k=0}^{\infty}\rho^{k}=\frac{1}{1-\rho}=l $, then we have that

"$\forall \frac{\epsilon}{2}>0 \text{ }\exists \text{ } N'_{\frac{\epsilon}{2}}=N'(\frac{\epsilon}{2}) \in \mathbb{N}$ s.t. $\forall n \geq N'_{\frac{\epsilon}{2}}$ , $ \left | \sum_{k=0}^{n} \rho^{k}-l \right | < \frac{\epsilon}{2} $"

Let $ \epsilon>0$, I choose $N_{\frac{\epsilon}{2}}=N'_{\frac{\epsilon}{2}}$ . We want to show that $N_{\frac{\epsilon}{2}}$ works.

$\text{ if } n,m\geq N_{\frac{\epsilon}{2}} \Rightarrow$

$\left | X_{n}-X_{m} \right |=\left | \sum_{k=0}^{n} \rho^{k}-\sum_{k=0}^{m} \rho^{k}\right |< \left | \sum_{k=0}^{n} \rho^{k}-l+l-\sum_{k=0}^{m} \rho^{k}\right | \leq \left | \sum_{k=0}^{n} \rho^{k}-l\right |+\left | \sum_{k=0}^{m} \rho^{k}-l \right |<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

$\therefore N_{\frac{\epsilon}{2}} $ works.

Could you give me any suggestions for the first sequence? Is the second demonstration correct?

Thanks in advance for your help.

Answer 1

For the first we have that $\frac {1}{k!}<\frac {1}{k^2}$ for $k\geq 4$ and thus $\sum_{k=n+1}^{m} \frac {1}{k!}\leq \sum_{k=n+1}^{m} \frac {1}{k^2}$ and $\sum_{k=n+1}^{m} \frac {1}{k^2}$ converges and so the sequence $x_m=\sum_{k=n+1}^{m} \frac {1}{k^2}$ is Cauchy, and thus $y_m=\sum_{k=n+1}^{m} \frac {1}{k!}$ is upper bounded by a Cauchy sequence,so $x_m$ is Cauchy(do your thing with $ε>0$).

For the second well done. But these sums converge and so there are Cauchy.You don't have to show that they are Cauchy with definition,unless you've benn asked too:)

Answer 2

For the first series: $X_{n} = \sum^{n}_{k=0} \frac{1}{k!} $ you could use the following approximation: $$ \left| X_{n} - X_{m} \right| = \left| \sum^{n}_{k=0} \frac{1}{k!} - \sum^{m}_{k=0} \frac{1}{k!} \right| = \left| \sum^{m}_{k=n+1} \frac{1}{k!} \right| \leq \left| \sum^{m}_{k=n+1} \frac{1}{e^{k}} \right| $$ for a all value of $n$ and $m$ greater than $ n_{0}$ (say $n_{0} =10$, i'm lazy and this seams a good enough aproximation ) . Now this could be written as $$ \left| \sum^{m}_{k=n+1} \frac{1}{e^{k}} \right| = \left| \sum^{m}_{k=n+1} e^{-k} \right| = \left| e^{-(n+1)} \sum^{m-(n+1)}_{k=0} e^{-k} \right| $$ Now you can see there first term will drop to infinity and the second series is finite and bound above by $$ \lim_{n \to \infty} \sum^{m}_{k=n+1} e^{-k} = \frac{1}{1-e^{-1}} $$ This leads to the final expression $$ \left| X_{n} - X_{m} \right| = \leq \left| \sum^{m}_{k=n+1} \frac{1}{e^{k}} \right| \leq \frac{e^{-(n+1)}}{1-e^{-1}} $$ From here on it isn't difficult to produce an $\epsilon - \delta$ proof. i.e. Take $\epsilon > 0$ random and $n$ so that $$ n > \max \left( n_{0},\log(\epsilon(1-e^{-1}))-1 \right) $$ And so on. Hope this helps.