<s>[INST] Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + ... + \frac{v^2}{v^3+v} = \frac{\frac{1}{v}}{1 + \frac{1}{v^3}} + \frac{\frac{1}{v}}{1 + \frac{2}{v^3}} + ... + \frac{\frac{1}{v}}{1+\frac{1}{v^2}} \rightarrow 0 $$
So, the limit is not equal to 1.
Where is my mistake ?
 [/INST] Although each term approaches zero, there are an infinite number of terms as indicated by there being exactly $v$ terms in the addition. The denominator of each term approaches $v^3$ as $v \to \infty$ as $v^3$ is much greater than $v$ or any value less than $v$. This means that the limit can simply be thought of as the following:
$$\lim_{v\to\infty} v\times \frac{v^2}{v^3}=1$$
</s>