0
$\begingroup$

I am an AP Calc BC student who has just recently been introduced to sequences and series.

The proof from the checked answer to this question makes sense:

If a series converges, then the sequence of terms converges to $0$.

However, it's still not clicking intuitively, and I'd really like an intuitive understanding of why this is so.

Why does this make intuitive sense?

$\endgroup$
  • $\begingroup$ Intuitively, $\sum_{k=0}^\infty a_k$ converges to some limit $\Lambda$ means once $N$ is large enough, all partial sums $\sum_{k=0}^n a_k$ for $n \ge N$ are very close to $\Lambda$. Now $a_n = (\sum_{k=0}^n a_k ) - (\sum_{k=0}^{n-1} a_k)$ is the difference of two numbers. If both numbers are very close to same number, their difference has to be very small. $\endgroup$ – achille hui Nov 7 '16 at 3:54
1
$\begingroup$

If you're looking for an intuitive, non-technical reason, here you go...

Think of it like walking along the number line. Each term in the series tells you how big of a step to take in the positive / negative direction. If a series converges, then your steps will pull you into one specific value, to which you get "infinitely close." The only way you can get close to this value after taking an arbitrarily large number of steps is if your individual steps get really small.

$\endgroup$
1
$\begingroup$

Let $S_n$ denote the partial sum of the convergent series $\sum_{n=1}^\infty a_n$.

So $S_n$ converges to some $l$ say.

Then $a_n=S_{n+1}-S_n\implies \lim_{n\to \infty } a_n=\lim_{n\to \infty }(S_{n+1}-S_n)=l-l=0$

$\endgroup$
  • $\begingroup$ eh this is still pretty much the proof that i linked and i get this but i still don't "get" it intuitively $\endgroup$ – Horse Nov 7 '16 at 3:45
1
$\begingroup$

An intuitive explanation:

Let $\epsilon > 0$ and suppose $\sum a_i = L$.

Let's suppose that the limit of the sequence was not $0$. Then, for arbitrarily large $N$, there would be some $a_i$, $i > N$, satisfying $|a_i| > \epsilon$. In fact, it's easy to see that there's a countably infinite number of such $a_i$ in the sequence.

Now, say we have $\epsilon_0 = \frac{\epsilon}{10^{10}}$. Whenever the sequence of partial sums reaches an $a_i$ satisfying the above requirement, the value of the sum will obviously be "pushed out" of the interval $(L - \epsilon_0, L + \epsilon_0)$. This "pushing out" will occur regardless of how far you get in the sequence, hence the series cannot converge.

$\endgroup$

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.