Sequences of positive real number Suppose  that $(a_{n})_{n=1}^\infty$ and $(b_{n})_{n=1}^\infty$ are sequences of positive real number, and that $$\limsup\frac{a_{n}}{b_{n}} &lt; \infty\;.$$ Prove that there is a constant $M$ so that $a_{n}\le Mb_{n}$ for all $n\ge 1$.
Note that $a_n\leq Mb_n$ is the same as $\dfrac{a_n}{b_n}\leq M$, and that $\left(\dfrac{a_n}{b_n}\right)$ is just another sequence of positive numbers.  So let's simplify this to a more general (but equivalent) statement:


  If $(c_n)$ is a sequence of positive numbers and $\limsup\limits_{n\to\infty}c_n&lt;\infty$, then there is a positive constant $M$ such that $c_n\leq M$ for all $n\geq 1$. 


It helps to give a name to the limsup, say $\limsup\limits_{n\to\infty}c_n = A$.  By definition, $\limsup\limits_{n\to\infty} c_n = \inf\limits_{n\geq 1}\sup\limits_{k\geq n}c_k = A$.  By definition of inf, this means that if we take any number $A'$ larger than $A$, then there will exist an $n_0$ such that $\sup\limits_{k\geq n_0}c_k&lt;A'$.  For example, we can take $A'=A+1$, and there exists $n_0$ such that $\sup\limits_{k\geq n_0}c_k&lt;A+1$. This gives us an upper bound for all of the terms $c_n$ with $n\geq n_0$.  For the rest, note that there are only finitely many so it is possible to find a bound for them, and combine the bounds to find $M$.

To summarize the $2$ main ingredients:


Every number larger than the limsup is eventually an upper bound for the terms in the sequence.
A sequence is bounded if it is eventually bounded.
Hints: Suppose

$$\lim_{n\to\infty}\frac{a_n}{b_n}=L\implies \exists\,K\in\Bbb N\;\;s.t.\;\;n&gt;K\implies1&gt;\left|\frac{a_n}{b_n}-L\right|\ge\left|\frac{a_n}{b_n}\right|-L$$

Well, what happens in you choose $\;M:=L+1\;$ ...?

Note: above, one has to assume in fact that the first equality on the left is for a subsequence $\,\frac{a_{n_k}}{b_{n_k}}\,$ of $\,\frac{a_n}{b_n}\,$ , but this is easy to fix...