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$\begingroup$ Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series? $\endgroup$ –
The Taylor expansion itself can be derived from mean value theorems which themselves are valid over the entire domain of the function. Then why doesn't the Taylor series converge over the entire domain? I understand the part about the convergence of infinite series and the various tests. But I seem to be missing something very fundamental here..
4. No, a bounded series does not necessarily converge. Consider the series $\displaystyle \sum (-1)^n $ (heavily related to Henning's example). It will forever oscillate between 0 and 1 (or -1 and 0, depending on the indices). But if the partial sums are bounded and monotonic, then it does converge.
The answer dealt with the series $\sum \frac {1} {n}$. It turns out that for any positive $\epsilon$, the series $\sum \frac {1} {n^ {1+\epsilon}}$ converges. We can take for example $\epsilon=0.0001$. So one can say that $\sum \frac {1} {n}$ diverges extremely reluctantly, and that close neighbours converge. Share.
Side fact: the series I wrote down at the start has the bonus property that each term in the sequence is larger than the corresponding term of the sequence. $$1+\frac12+\frac13+\frac14+\frac15+\cdots$$ which is also known as the harmonic series and is the most famous divergent series.
According to your logic, since the sine function is periodic, this sum can't converge. But it does converge. How does one show this? Perhaps it's easiest to show that the partial sums are Cauchy. This is true because the tail of the k k th partial sum is contained within a distance at most ∑∞ k+1 1 n2 ∑ k + 1 ∞ 1 n 2 away, which ...
On the other hand, the series $\sum_{n=0}^\infty \frac{1}{n^2}$ does converge, to $\pi^2/6$, in fact. We can show that it converges using various theorems, one of them includes the integral test. To find the value of the sum requires more work. So at the end of the day, we have to use specific tools to show specific series either converge or ...
$\begingroup$ I think this is an interesting answer but you should use \frac{a}{b} (between dollar signs, of course) to express a fraction instead of a/b, and also use double line space and double dollar sign to center and make things bigger and clear, for example compare: $\sum_{n=1}^\infty n!/n^n\,$ with $$\sum_{n=1}^\infty\frac{n!}{n^n}$$ The first one is with one sign dollar to both sides ...
Here is another possible answer. We will derive the asymptotic formula of the partial sum $\sum_{1< n\leqslant x}\frac{1}{n\log n}$ to show that this series diverges.