# 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 ?

• Does $v$ approach $\infty$? – Peter Foreman Feb 16 at 11:32
• It is a sequence so we want to test the case where v approch $\inf$ – Dimitris Dimitriadis Feb 16 at 11:34
• Hint: each summand is of order ${1 \over v}$, and you have $v$ summands. – lisyarus Feb 16 at 11:35

## 3 Answers

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 $$0$$.

• It am a bit confused. I understand that there are n terms but each term is equal to zero. So $n*0 = 0$ . Why this is wrong ? – Dimitris Dimitriadis Feb 16 at 11:48
• MMMmaybe because we replace $n * \frac{1}{n}$ before finding the lim ? – Dimitris Dimitriadis Feb 16 at 11:51
• We have more and more terms and, on the other hand, the terms are getting smaller. And these two factors compensate each other. You cannot deal with them separately. – José Carlos Santos Feb 16 at 11:54
• @DimitrisDimitriadis, the precise point where you take the limit is important. A rather artificial example for the sake of illustration: Surely you agree $\lim\limits_{n \to \infty} 1 = 1$, if you write $\lim\limits_{n \to \infty} \frac{n}{n}$ the limit is still one, however what happens if you split the fraction an take the limit? $(\lim\limits_{n \to \infty} \frac{1}{n}) * (\lim\limits_{n \to \infty} n)$ ends in a mess. It is vey important in maths, when you take the limit of a function, because as seen: $\lim f(a_n) = f( \lim a_n)$ does in general not have to be true. – Imago Feb 16 at 12:34
• @JoséCarlosSantos your last comment seems like a theorem and it is very rational. Could you share a source about this ? – Dimitris Dimitriadis Feb 16 at 12:48

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 $$\lim c_v = \lim b_v = 1$$

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: $$\lim_{v\to\infty} v\times \frac{v^2}{v^3}=1$$