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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 ?

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    $\begingroup$ Does $v$ approach $\infty$? $\endgroup$ – Peter Foreman Feb 16 at 11:32
  • $\begingroup$ It is a sequence so we want to test the case where v approch $\inf$ $\endgroup$ – Dimitris Dimitriadis Feb 16 at 11:34
  • $\begingroup$ Hint: each summand is of order ${1 \over v}$, and you have $v$ summands. $\endgroup$ – lisyarus Feb 16 at 11:35
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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 $0$.

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  • $\begingroup$ 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 ? $\endgroup$ – Dimitris Dimitriadis Feb 16 at 11:48
  • $\begingroup$ MMMmaybe because we replace $ n * \frac{1}{n} $ before finding the lim ? $\endgroup$ – Dimitris Dimitriadis Feb 16 at 11:51
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    $\begingroup$ 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. $\endgroup$ – José Carlos Santos Feb 16 at 11:54
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    $\begingroup$ @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. $\endgroup$ – Imago Feb 16 at 12:34
  • $\begingroup$ @JoséCarlosSantos your last comment seems like a theorem and it is very rational. Could you share a source about this ? $\endgroup$ – Dimitris Dimitriadis Feb 16 at 12:48
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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 $$\lim c_v = \lim b_v = 1$$

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

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