What's the largest term in a converging series? Although it's quite a trivial question but is it always that the very first term in a converging series is the largest term of all the terms in that series ?
Since if $\sum A_n$ converges, that is if $\sum A_n = a_0 + a_1 + a_2 +......$ converges then probably $a_n$ must be growing smaller and smaller as $n$ increases . If not then what are the exceptions ?
 A: For a series $\sum a_n$ to converge, you need that $a_n\to 0$ as $n\to\infty$. But that does not mean you need to have $a_n>a_{n+1}$ for all $n$. For instance, the series
$$\sum_{n=1}^{\infty}\frac{1}{n^2}\frac{\sin(n)}{n}$$
converges, but have a look at the first 20 values:

Edit: In fact, there is no need of an alternating series. For instance,
$$\sum_{n=1}^{\infty}\frac{1}{n^2}\frac{k+\sin(n)}{n},\quad\mbox{with }k\ge 1.$$
Take a look at the case $k=1$.

A: A very simple counterexample is the series $$0+0+…+1+0+0+…$$ That is, for each $n$, we have the sum $1 = \sum_{k=1}^∞ a^{(n)}_k$ of the sequence $a^{(n)}_k = \begin{cases}1 & k=n \\ 0 & k≠n\end{cases}$, which can be made to have the $k_0$th term $a^{(n)}_{k_0}$ biggest by choosing $n=k_0$.
A: All you can say is that the largest term occurs in a finite place, not necessarily the first.
Think of a convergent series, now add an extra term, larger than all terms in the original sequence and make it your new $a_{100}$, the new series will converge to the original limit plus this new term, but the largest term will be in the $100^{th}$ place.
A: The exponential series
$$\exp(z):=\sum_{k=0}^\infty{1\over k!}\>z^k\qquad(z\in {\mathbb C})$$
provides an infinity of examples. It converges for any given $z\in{\mathbb C}$, but the position of its largest term (in absolute value) depends heavily on the chosen $z$: Put
$$a_k:=\left|{z^k\over k!}\right|\ .$$
Then $${a_k\over a_{k-1}}={|z|\over k}$$
is $>1$ as long as $k<|z|$, and $<1$ when $k>|z|$. Therefore the $a_k$ are increasing at the start, reach a maximum when $k$ is is one of $\bigl\lfloor |z|\bigr\rfloor$ and $\bigl\lceil |z|\bigr\rceil$, and then decrease monotonically to $0$.
