# Thread: Two proofs regarding limits

1. ## Two proofs regarding limits

I just want to make sure I have done these proofs correctly.

1) Let $\displaystyle (t_{n})$ be a bounded sequence such that there exists $\displaystyle M$ such that $\displaystyle |t_{n}| \le M$ for all $\displaystyle n$, and let $\displaystyle (s_{n})$ be a sequence such that $\displaystyle lim\, s_{n} = 0$. Prove $\displaystyle lim\, (s_{n}t_{n}) = 0$.

PROOF:

Assume the given conditions above. This means that there exists $\displaystyle N$ such that, if $\displaystyle \epsilon > 0$ and $\displaystyle n > N$, then $\displaystyle |s_{n}| < \epsilon$. In particular, there exists $\displaystyle N_{1}$ such that if $\displaystyle \epsilon > 0$ and $\displaystyle n > N_{1}$, then $\displaystyle |s_{n}| < \frac{\epsilon}{M}$. Hence, $\displaystyle |s_{n}t_{n}| = |s_{n}||t_{n}| < \frac{\epsilon}{M}|t_{n}| \le \frac{\epsilon}{M}M = \epsilon$, given that $\displaystyle \epsilon > 0$ and $\displaystyle n > N_{1}$. And so, $\displaystyle lim\,(s_{n}t_{n}) = 0$. Q.E.D.

2)Suppose that $\displaystyle (s_{n})$ and $\displaystyle t_{n}$ are sequences such that $\displaystyle |s_{n}| \le t_{n}$ for all $\displaystyle n$ and $\displaystyle lim\, t_{n} = 0$. Prove $\displaystyle lim\, s_{n} = 0$.

PROOF:

Assume the given conditions above. This means there exists $\displaystyle N$ such that, if $\displaystyle \epsilon > 0$ and $\displaystyle n > N$, then $\displaystyle |t_{n}| < \epsilon$. Hence, $\displaystyle |s_{n}|\le t_{n} = |t_{n}| < \epsilon$, and so there exists $\displaystyle N$ such that, if $\displaystyle \epsilon > 0$ and $\displaystyle n > N$, then $\displaystyle |s_{n}| < \epsilon$. And so, $\displaystyle lim\, s_{n} = 0$. Q.E.D.

Any feedback would be appreciated, thanks.

2. Originally Posted by Pinkk
I just want to make sure I have done these proofs correctly.

1) Let $\displaystyle (t_{n})$ be a bounded sequence such that there exists $\displaystyle M$ such that $\displaystyle |t_{n}| \le M$ for all $\displaystyle n$, and let $\displaystyle (s_{n})$ be a sequence such that $\displaystyle lim\, s_{n} = 0$. Prove $\displaystyle lim\, (s_{n}t_{n}) = 0$.

PROOF:

Assume the given conditions above. This means that there exists $\displaystyle N$ such that, if $\displaystyle \epsilon > 0$ and $\displaystyle n > N$, then $\displaystyle |s_{n}| < \epsilon$. In particular, there exists $\displaystyle N_{1}$ such that if $\displaystyle \epsilon > 0$ and $\displaystyle n > N_{1}$, then $\displaystyle |s_{n}| < \frac{\epsilon}{M}$. Hence, $\displaystyle |s_{n}t_{n}| = |s_{n}||t_{n}| < \frac{\epsilon}{M}t_{n} \le \frac{\epsilon}{M}M = \epsilon$, given that $\displaystyle \epsilon > 0$ and $\displaystyle n > N_{1}$. And so, $\displaystyle lim\,(s_{n}t_{n}) = 0$. Q.E.D.

2)Suppose that $\displaystyle (s_{n})$ and $\displaystyle t_{n}$ are sequences such that $\displaystyle |s_{n}| \le t_{n}$ for all $\displaystyle n$ and $\displaystyle lim\, t_{n} = 0$. Prove $\displaystyle lim\, s_{n} = 0$.

PROOF:

Assume the given conditions above. This means there exists $\displaystyle N$ such that, if $\displaystyle \epsilon > 0$ and $\displaystyle n > N$, then $\displaystyle |t_{n}| < \epsilon$. Hence, $\displaystyle |s_{n}|\le t_{n} = |t_{n}| < \epsilon$, and so there exists $\displaystyle N$ such that, if $\displaystyle \epsilon > 0$ and $\displaystyle n > N$, then $\displaystyle |s_{n}| < \epsilon$. And so, $\displaystyle lim\, s_{n} = 0$. Q.E.D.

Any feedback would be appreciated, thanks.