# Thread:

1. ## Some questions on proving convergence tests

I was reading proofs on some convergence tests, but I'm stuck on a few parts here:

1. Prove every cauchy sequence is bounded.

Proof: Suppose the sequence {An} is cauchy.
Let E = 1, pick N in Natural Numbers such that n >= N and m >= N

We have |An - Am| < 1, further, we have |An-AN| < 1.

This is the part that I don't understand, how do we know that|An - AN| < 1?

2. For a number r such that |r| < 1, then the series sum {from k=0} r^k = 1/(1-r).

Proof: sum {from k=0 to n} r^k = 1 + r + r^2 + r^3 + . . . + r^n = (1 - r^{n+1}) / (1 - r)

Now, how do we know that equals?

Thank you, y'all!

KK

2. Theorem: If $(s_n)$ is a Weierstrauss sequence then $s_n$ is bounded.

Proof: We now that for any $\epsilon$ we can choose $N\in \mathbb{N}$ so that $n,m>N\implies |s_n-s_m|<\epsilon$. So choose $\epsilon = 1$ and we have $|s_n - s_m|<1$. That means
$|s_n|-|s_m|\leq|s_n - s_m|<1 \mbox{ thus }|s_n|< 1+ |s_{N+1}|$ (by choose $m=N+1$).
Now the above inequality if true for only $n>N$.
So,
$|s_n|<1+|s_{N+1}|$ for $n>N$.
But what if $n=1,2,..,N$?
No problem, let,
$M= \max \{ s_1,s_2,...,s_n , 1+|s_{N+1}| \}$.
And so,
$|s_n|\leq M \mbox{ for all } n\in \mathbb{N}$.

3. Originally Posted by tttcomrader
2. For a number r such that |r| < 1, then the series sum {from k=0} r^k = 1/(1-r).
If $|r|<1$ then,
$1+r+r^2+...+r^n = \frac{1-r^{n+1}}{1-r}\to \frac{1}{1-r} \mbox{ as }n\to \infty \mbox{ since }|r|<1$

4. I understand the first proof now, thank you.

But in the second one, why does

$<br /> 1+r+r^2+...+r^n = \frac{1-r^{n+1}}{1-r}<br />$

true?

Other than that, I understand the rest, thank you!

5. Geometric series formula + the following theorem:
$\lim \ a^n = 0 \mbox{ if } |a|<1$.
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