Question 1: Let $(X, d)$ be a metric space and $(x_n)$ be a sequence in $X$. Let $(x_n)$ be a Cauchy sequence, prove that if $(x_n)$ has a cluster point $z$, then $(x_n) \to z$.

My attempt: Suppose that $z$ is a cluster point of $(x_n)$, and construct a subsequence $x_{n_k}$ with limit $z$.

Choose $x_{n_1} \in B(z,1)$, $B(z,1/2)$ contains infinitely many elements of $(x_n)$

Choose $x_{n_2} \in B(z,1/2)$ such that $n_2 > n_1$

..... ..... .....

Choose $x_{n_k} \in B(z,1/k)$ so that $n_k > n_{k-1} > \cdots$

By the choice we did, $d(x_{n_k},z) < 1/k$ and we know that $1/k \to 0$ then, $d(x_{n_k},z) \to 0$. Hence, $x_{n_k} \to z$.

Now how can I show that $x_n$ also converges to $z$?

Question 2: Let $a_n > 0$, $b_n > 0$, $n=1,2,\dots$. Prove that if $\frac{a_n}{b_n} \to l \neq 0$, then the series $\sum_{n=1}^{\infty} a_n$ and $\sum _{n=1}^{\infty} b_n$ converge or diverge simultaneously.

My attempt: If $\frac{a_n}{b_n} \to l \neq 0$ then we can also say that if $\frac{a_n}{b_n}$ is closed by $l$.

If $l>0$, then there are positive numbers $c$ and $d$ with $0<c\leq l \leq d$ with $c \leq \frac{a_n}{b_n} \leq d$. For $b_n$>0, $cb_n \leq a_n \leq db_n$.

Now if $\sum_{n=1}^{\infty} b_n$ converges, then so does $\sum_{n=1}^{\infty} db_n$. From the inequality $\frac{a_n}{b_n} \leq d$, $\sum_{n=1}^{\infty} a_n$ also converges. However if $\sum_{n=1}^{\infty} b_n$ diverges, then so does $\sum_{n=1}^{\infty } cb_n$.

From $a_n \leq db_n$ we see that if $\sum_{n=1}^{\infty} a_n$ diverges then $\sum_{n=1}^{\infty} \frac{a_n}{d}$ also diverges and in this case $\sum_{n=1}^{\infty} b_n$ diverges too. With similar argument we can see that if $\sum_{n=1}^{\infty} a_n$ converges then $\sum_{n=1}^{\infty} \frac{a_n }{c}$ also converges and from the left inequality ($b_n < \frac{a_n}{c}$) $\sum_{n=1}^{\infty} b_n$ converges too.

Is this proof right and enough?

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These are two entirely different questions in different topics. Please delete one question (edit) and re-post that question as a separate question. –  amWhy Nov 26 '12 at 0:35
Excuse me, that's THREE separate questions in two different topics, in ONE post? –  amWhy Nov 26 '12 at 1:00
@amWhy you are right. I was a little bit panicked to get a fast answer that is why I did such a thing. I have already got answers for the first two questions. So I only deleted the third question and will post it as a separate question. –  Amadeus Bachmann Nov 26 '12 at 1:17
"if $\frac{a_n}{b_n}$ is closed by $l$" (as in line 1 of your attempt on question 2) does not mean anything, and it is not even a statement. Maybe you meant to say that if $a_n/b_n\to l$ then $a_n/b_n$ is close to $l$? The problem is that even though this informal statement is true (for large $n$, anyway), it is useless, precisely because it is informal. –  Andres Caicedo Nov 26 '12 at 1:49
@AndresCaicedo yes I meant to say "it is close to $l$" but now I am aware that that statement was useless. –  Amadeus Bachmann Nov 26 '12 at 2:00

For the series question use the definition of the limit of a sequence. Let $\epsilon=\frac{l}{2}$, then

$$\left| \frac{a_n}{b_n} - l\right| < \frac{l}{2}\quad \mathrm{when}\> n>N$$

$$\implies \frac{l}{2}<\frac{a_n}{b_n}<\frac{3l}{2} \quad \mathrm{when}\> n>N$$

$$\implies \frac{l}{2} b_n< a_n <\frac{3l}{2}b_n \quad \mathrm{when}\> n>N$$

Now, if $\sum b_n$ converges, so does $\sum \frac{3l}{2}b_n$. Then the right half of the above inequality shows that $\sum_{n=N}^{\infty}a_n$ converges by the comparison test. It follows that $\sum_{1}^{\infty}a_n$ converges. I think you can finish the proof now.

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Thank you so much! –  Amadeus Bachmann Nov 26 '12 at 1:11
@Zxy: You are welcome. –  Mhenni Benghorbal Nov 26 '12 at 1:14
@MehenniBenghorbal Oh it just came to my mind: is it OK to directly say that let $\epsilon=\frac{l}{2}$ for a formal proof? I know it does not change anything as long as $\epsilon$>0. But I had a feeling that my professor would not be happy if I directly say let $\epsilon=\frac{l}{2}$. = ) –  Amadeus Bachmann Nov 26 '12 at 2:08

I'll answer the first question. You have a Cauchy sequence that has a convergent subsequence, and you'd like to show the sequence converges. So, let $\epsilon>0$. Since $(x_n)$ is Cauchy, we can choose $N_1$ so that for every $n,m\geq N_1$ we have $d(x_n,x_m)<\frac{\epsilon}{2}$. Then, since the subsequence $(x_{nk})$ converges to $z$, we can choose $N_2$ so that for every $k\geq N_2$, $d(x_{nk},z)<\frac{\epsilon}{2}$. Then, set $N=\max\{N_1,N_2\}$ so that for every $n>N$, we have by the triangle inequality

$$d(x_n,z)\leq d(x_n,x_{nn})+d(x_{nn},z)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

Thus $(x_n)$ indeed converges to $z$.

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I am very grateful for your help! –  Amadeus Bachmann Nov 26 '12 at 1:14