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If the limit of sequence is zero then so is the limit of sequence multiplied by a bounded one

Solution: Let $\epsilon>0$. Since $\lim s_n=0$, there exists $N>0$ such that
|s_n-0|=s_n < \frac{\epsilon}{M+1}
$$ for all $n>N$. Therefore $n>N$ implies that
$$ for all $n>N$ as desired. Thus $\lim (s_nt_n)=0$.

Here we use $\dfrac{\epsilon}{M+1}$ instead of $\dfrac{\epsilon}{M}$ because $M$ may be zero. With this modification, we do not need to consider the case where $M=0$ separately.

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