8
$\begingroup$

I understand that when a series diverges, y doesn't approach 0 when x approaches infinity, and converging series do. But what does this say? I just want to understand some applications.

NOTE: I'm still on integration by parts, so try to explain in simple terms. This site scares me sometimes... xD

Thanks!

EDIT: I got the x-axis and y-axis mixed up when I wrote it. I do understand what they actually mean. Sorry!

$\endgroup$
0
9
$\begingroup$

When a sequence converges, that means that as you get further and further along the sequence, the terms get closer and closer to a specific limit (usually a real number).

A series is a sequence of sums. So for a series to converge, these sums have to get closer and closer to a specific limit as we add more and more terms up to infinity. For example, to see if the infinite series $$\sum_{k=1}^{\infty} (\frac{1}{n^2})$$ converges, we analyse whether the sequence of sums given by$$1, 1+ \frac{1}{2^2}, 1+ \frac{1}{2^2} + \frac{1}{3^2}, \dots$$ gets closer and closer to a limit as we go further along this sequence (it turns out the limit is actually $\frac{\pi^2}{6}$, try adding up a lot of terms on a calculator to see that the series tends to this limit).

Something diverges when it doesn't converge. Notoriously the series $$\sum_{k=1}^{\infty} (\frac{1}{n})$$ actually diverges, as an example.

$\endgroup$
3
$\begingroup$

Very informally, a sequence converges when there is a point, called the "limit", and the terms in the sequence get and remain as close as you want to this limit.

Consider it a game: The "other" specifies a distance. If you can specify a location in the sequence such that all items in the sequence beyond this point are within that distance of the limit, and do this for any distance, you win, and the sequence converges.

If there is some distance such that no matter how far you go out in the sequence, you can find two items that are at least that distance apart, the sequence does not converge, and is said to "diverge".

This divergence can occur in more than one way.

For example, if the n-th term of the sequence is $n$, the sequence diverges to infinity.

If the n-th term is $(-1)^n$, the sequence oscillates. Note that since $|a_{2n}-a_{2n+1}| = 2 $, we can choose any distance $d$ less that 2 and, for any integer $m$, find two terms beyond the m-th which differ by more that $d$. More explicitly, if we choose $a_{m+1}$ and $a_{m+2}$, then $|a_{m+1}-a_{m+2}| =2 > d $ for any $m$ and any $d < 2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.