Help please I’ll mark you brainliest

Mathematics

1 answer:

prohojiy [21]2 years ago

3 0

Subtract any number from the next one.
-1 - (-6)
4 - (-1)
9 - 4
14 - 9

All of the above subtractions give the same answer.
That answer is the common difference.

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Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .