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 ?
 A: Consider the sequence$$1,\frac12+\frac12,\frac13+\frac13+\frac13,\ldots,\overbrace{\frac1n+\cdots+\frac1n}^{n\text{ times}},\ldots$$Its limit is $1$, right?! However, by your argument, the limit should be $0$.
Yes, $\lim_{n\to\infty}\frac1n=0$, but, on the other hand, you have $n$ terms. So, the fact that the limit of each term is $0$ does not imply that the limit of the whole sequence is $0$.
A: I think that its good idea to use the sandwich theorem.
Let $a_v $ the sequence that you posted. Then
$$a_v \leq \frac{v^2}{v^3 + v} + \cdots + \frac{v^2}{v^3 +v} = b_v $$
And
$$b_v = v\left(\frac{v^2}{v^3+v}\right) \to 1$$
On the other hand you must use
$$c_v = \frac{v^2}{v^3 +1} + \cdots + \frac{v^2}{v^3 +1} \leq a_v $$
And..
$$c_v = v\left(\frac{v^2}{v^3+1}\right) \to 1$$
And 
$$c_v \leq a_v \leq b_v $$
With $$\lim c_v = \lim b_v = 1$$
A: 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:
$$\lim_{v\to\infty} v\times \frac{v^2}{v^3}=1$$
