Two proofs regarding limits

• Feb 3rd 2010, 06:57 PM
Pinkk
Two proofs regarding limits
I just want to make sure I have done these proofs correctly.

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

PROOF:

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

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

PROOF:

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

Any feedback would be appreciated, thanks.
• Feb 3rd 2010, 06:59 PM
Drexel28
Quote:

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

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

PROOF:

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

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

PROOF:

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

Any feedback would be appreciated, thanks.

(Yes)