# Thread: Sequences and series

1. ## Sequences and series

Consider a convergent sequence $(x_{n})_{n}$ such that $limx_{n} = x$.

(a) Let $(y_{n})_{n}$ be built such that $y_{n} = x_{n+100}$ , for all $n\in\mathbb{N}$. Is $(y_{n})_{n}$ convergent? If yes, what is the
limit?
(b) Let $(z_{n})_{n}$ be now the sequence $\{0, \pi\\, e, -\sqrt{2}, ln(2), x_{1}, x_{2} , x_{3} , . . . , x-{n} , . . . \}$. In other words, we added 5 extra terms at the beginning of $(x_{n})_{n}$ and we called this new sequence $(z_{n})_{n}$ . Is $(z_{n})_{n}$convergent?
If yes, what is the limit?
(c) Consider now a convergent series $\sum_{
n=1}a_{n}$
. Fix some natural number $K > 1$. Is $\sum_{n=K}a_{n}$ convergent?

(a) Assume $(x_{n})_{n}$ converges to x and let $y_{n} = x_{n+10}0$. Since $(x_{n})_{n}$ is convergent, $(x_{n})_{n}$ has convergent subsequences. Moreover, these subsequence congerge to the same limit as $(x_{n})_{n}$, $x$. Cleary, $y_{n} = x_{n+100}$ is a subsequence of $(x_{n})_{n}$. Thus, $(y_{n})_{n}$ converges to $x$.

Is it enough just to use a theorem to prove this one?

(b) Can I just note that the first 5 terms are constants and pull them out of the limit and conclude the sequence still converges to $x$? I am not sure how to formalize this. Do I need to use the definition of convergence or the cauchy definition to prove this?

(c) Not totaly sure how to approach this one. I would think that is iss convergent, because, even though K is some number latter on in the sequence, there will still be a rank for which the terms in the sequence are stuck in some epsilon neighborhood.

Can anyone help me with some formal proofs?

Thank you

2. Originally Posted by Danneedshelp
Consider a convergent sequence $(x_{n})_{n}$ such that $limx_{n} = x$.

(a) Let $(y_{n})_{n}$ be built such that $y_{n} = x_{n}+100$ , for all $n\in\mathbb{N}$. Is $(y_{n})_{n}$convergent? If yes, what is the limit?
(b) Let $(z_{n})_{n}$ be now the sequence $\{0, \pi\\, e, -\sqrt{2}, ln(2), x_{1}, x_{2} , x_{3} , . . . , x-{n} , . . . \}$. In other words, we added 5 extra terms at the beginning of $(x_{n})_{n}$ and we called this new sequence $(z_{n})_{n}$ . Is $(z_{n})_{n}$convergent?
If yes, what is the limit?
(c) Consider now a convergent series $\sum_{
n=1}a_{n}$
. Fix some natural number $K > 1$. Is $\sum_{n=K}a_{n}$ convergent?

(a) Assume converges to x and let $y_{n} = x_{n}+100$. Since $(x_{n})_{n}$ is convergent, $(x_{n})_{n}$ has convergent subsequences. Moreover, these subsequence congerge to the same limit as $(x_{n})_{n}$, $x$. Cleary, $y_{n} = x_{n}+100$ is a subsequence of $(x_{n})_{n}$. Thus, $(y_{n})_{n}$ converges to $x$.

Is it enough just to use a theorem to prove this one?

(b) Can I just note that the first 5 terms are constants and pull them out of the limit and conclude the sequence still converges to $x$? I am not sure how to formalize this. Do I need to use the definition of convergence or the cauchy definition to prove this?

(c) Not totaly sure how to approach this one. I would think that is iss convergent, because, even though K is some number latter on in the sequence, there will still be a rank for which the terms in the sequence are stuck in some epsilon neighborhood.

Can anyone help me with some formal proofs?

Thank you
(a) This sequence converges to $x+100$.

Proof: Since $\{x_n\}\to x$, then $\forall~\epsilon>0$, $\exists~N$ such that $n>N$ implies that $|x_n-x|<\epsilon$. Thus, $n>N$ also implies that $|(x_n+100)-(x+100)|=|x_n-x|<\epsilon$. So $\{y_n\}\to x+100$. $\square$

---------------

(b) This sequence converges to x.

Proof: Since $\{x_n\}\to x$, then $\forall~\epsilon>0$, $\exists~N$ such that $n>N$ implies that $|x_n-x|<\epsilon$. Thus, $n>N+5$ implies that $|z_n-z|<\epsilon$. So $\{z_n\}\to x$. $\square$

---------------

(c) It converges.

Proof: $\sum_{n=K}^{\infty}a_n=\sum_{n=1}^{\infty}a_n -\sum_{n=1}^{K-1}a_n$. Looking at the right side of the equation, we have a finite sum subtracted from a convergent series, so it is obvious that the right side is finite, and therefore $\sum_{n=K}^{\infty}a_n$ converges. $\square$

3. Originally Posted by redsoxfan325
(a) This sequence converges to $x+100$.

Proof: Since $\{x_n\}\to x$, then $\forall~\epsilon>0$, $\exists~N$ such that $n>N$ implies that $|x_n-x|<\epsilon$. Thus, $n>N$ also implies that $|(x_n+100)-(x+100)|=|x_n-x|<\epsilon$. So $\{y_n\}\to x+100$. $\square$

---------------

(b) This sequence converges to x.

Proof: Since $\{x_n\}\to x$, then $\forall~\epsilon>0$, $\exists~N$ such that $n>N$ implies that $|x_n-x|<\epsilon$. Thus, $n>N+5$ implies that $|z_n-z|<\epsilon$. So $\{z_n\}\to x$. $\square$

---------------

(c) It converges.

Proof: $\sum_{n=K}^{\infty}a_n=\sum_{n=1}^{\infty}a_n -\sum_{n=1}^{K-1}a_n$. Looking at the right side of the equation, we have a finite sum subtracted from a convergent series, so it is obvious that the right side is finite, and therefore $\sum_{n=K}^{\infty}a_n$ converges. $\square$
So, in general, if I have a new series $y_{n}=x_{n+c}$. I only have to show $|(x_{n}-c)-(x-c)|<\epsilon$ holds?

For part (b) did you mean, $|z_{n}-x|<\epsilon$?

4. Originally Posted by Danneedshelp
So, in general, if I have a new series $y_{n}=x_{n+c}$. I only have to show $|(x_{n}-c)-(x-c)|<\epsilon$ holds?

For part (b) did you mean, $|z_{n}-x|<\epsilon$?
Yes, for a series $y_n=x_n+c$ (is that what you meant?), you have that $|(x_n+c)-(x+c)|=|x_n-x|<\epsilon$ so $y_n\to x+c$.

And for part (b), yes, that's what I meant.

5. Originally Posted by redsoxfan325
Yes, for a series $y_n=x_n+c$ (is that what you meant?), you have that $|(x_n+c)-(x+c)|=|x_n-x|<\epsilon$ so $y_n\to x+c$.

And for part (b), yes, that's what I meant.
Oh boy, I wrote the first problem down wrong!

It should be...

(a) Let $(y_{n})_{n}$ be built such that $y_{n} = x_{n+100}$ , for all $n\in\mathbb{N}$. Is $(y_{n})_{n}$ convergent? If yes, what is the
limit?

Suppose $(x_{n})_{n}$ converges to $x$ and let $y_{n}=x_{n+100}$. Since $(x_{n})_{n}$ converges, there exists an $N\in\mathbb{N}$ such that $n\geq\\N$ implies $|x_{n}-x|<\epsilon$. Now, let $N$ be some natural number satisfying $N\geq\\n+100$ such that for all $n\geq\\N$ implies $|x_{n+100}-x|=|y_{n}-x|<\epsilon$. Therefore, $(y_{n})_{n}$ converges to x as well.

6. Yes. There is a theorem that states that a sequence $\{x_n\}$ converges to $x$ if and only if every (infinite) subsequence of $\{x_n\}$ also converges to $x$.