Boundedness of sequences In order to prove that the sequence $a_n$=($1^k+2^k+3^k+..n^k$)$/n!$ is bounded(k is fixed natural number).Is it enough to say that $a_n<nn^k/n!$.What bothers me is that following the definition of the boundedness of a sequence there must exist M such that $|a_n|<M$ but can we consider $M$ that is expressed in terms of index of $a_n$? As in example above can the bound be  $nn^k/n!$?
 A: $M$ should be a constant that does not depend on $n$. It's true that $a_n \le n^{k+1}/n!$, but that doesn't answer the question yet. If you can show that the sequence $\{n^{k+1}/n!\}_{n \ge 1}$ is bounded by some $M$ (that does not depend on $n$), then we have
$$0 \le a_n \le \frac{n^{k+1}}{n!} \le M,$$
which does prove that $\{a_n\}$ is bounded.
The more general fact that you are using is:

If $\{a_n\}$ and $\{b_n\}$ are nonnegative sequences with $\{b_n\}$ bounded and $a_n \le b_n$ for all $n$, then $\{a_n\}$ is also bounded.

A: You are correct to say that $|a_n| < n^{k+1} / n!$, but this does not show the function is bounded by some constant (i.e. not depending on $n$). To find such a bound, we can use the fact that there is $n_0$ such that $n^{k+1} / n! < 1$ for all $n \geq n_0$ (i.e. $n!$ eventually dominates $n^{k+1}$ for constant $k$). We can therefore let
$$
A = \bigcup_{i = 1}^{n_0} \{i^{k+1} / i!\} \cup \{1\}
$$
and take
$$
M = \max A.
$$
That is, we choose as our bound one of the first $n_0$ terms of the sequence or $1$, whichever is largest.
