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I got a question to prove, using the formal definition of sequence limits, that the limit of the series $A(n) = \frac{2n^2 + 3n}{5-3n^2}$ is $-\frac{2}{3}$.

Formal Definition of Series Limit

We call $x$ the limit of the sequence $a(n)$ if the following condition holds:

For each real number $\epsilon > 0$, there exists a natural number $N$ such that, for every natural number $n>N$, we have $\lvert a(n) - x\rvert < \epsilon$.

I tried to crack this all day, but I can't get past the inequality.

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    $\begingroup$ Factor out by $n^2$. $\endgroup$ – hamam_Abdallah Nov 2 '16 at 20:08
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    $\begingroup$ I think you mean sequence, not series. A series is when you're adding them all together. $\endgroup$ – Jam Nov 2 '16 at 20:11
  • $\begingroup$ Solve it for the corresponding real function f: x ____> (2x^2+3n)/(5-3x^2) using integration and differentiation. $\endgroup$ – Jacob Wakem Nov 2 '16 at 20:18
  • $\begingroup$ Consider the limit of the quotient of polynomials in its full generality. $\endgroup$ – Jacob Wakem Nov 2 '16 at 20:21
  • $\begingroup$ The numerator and denominator get arbitarily close in terms of ratio to 2n^2 and -3n^2 respectively. That is A(n) is approximated by (2n^2(1+e))/((-3n^2)(1-e)) $\endgroup$ – Jacob Wakem Nov 2 '16 at 20:26
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HINT: Divide both the nominator and numerator by $n^2$ and use the fact that $\lim_{n \to \infty} \frac{\text{const}}{n^2} = 0$


For the formal proof, choose $N = \max\{\frac {1}{\epsilon}, 2\}$

$$\left | \frac{2n^2 - 3n}{5-3n^2} + \frac 23 \right | = \left | \frac{6n^2 - 9n + 10 - 6n^2}{15-9n^2}\right | = \left | \frac{9n-10}{9n^2-15}\right | = \left | \frac{1 - \frac{10}{9n}}{n - \frac{15}{9n}}\right | < \left | \frac{1}{n - \frac{15}{9n}}\right | < \left | \frac{1}{n - 1}\right | < \frac{1}{N} \le \epsilon$$

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    $\begingroup$ Since the definition has $n\gt N$, I think you can get away with choosing $N=\max\{{1\over\epsilon},1\}$. That is, you only need $n\ge2$ for $n-{15\over9n}\gt n-1$. $\endgroup$ – Barry Cipra Nov 2 '16 at 20:28
  • $\begingroup$ @BarryCipra Yeah, you're right, although sometimes the inequality isn't strict depending from an author to author. $\endgroup$ – Stefan4024 Nov 2 '16 at 20:29
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    $\begingroup$ True, some authors use $n\ge N$. I was just going with what the OP said. $\endgroup$ – Barry Cipra Nov 2 '16 at 20:31
  • $\begingroup$ @BarryCipra Right. Anyway I guess that's going too much into details. After all we might not need the $1$ at the end. All we need is to define $N$ as the smallest integer greater than $\frac {1}{\epsilon}$. We're guaranteed $N \ge 1$. $\endgroup$ – Stefan4024 Nov 2 '16 at 20:33
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    $\begingroup$ You are quite right. The take-home message for the OP, I hope, is that you need to choose $N$ large enough so that all of the final four inequalities in the display are satisfied. $\endgroup$ – Barry Cipra Nov 2 '16 at 20:59
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The trick to solve a lot of such questions that go to $\infty$ is that $$\lim_{n\to\infty} \frac{1}{n} = 0$$

So, here, try to put the largest terms in the denominator. This can be done by dividing both the numerator and denominator by $n^2$ to reduce the expression to $$\lim_{n\to\infty} \frac{2+\frac{3}{n}}{\frac{5}{n}-3}$$

Now, both the terms $\frac{3}{n}$ and $\frac{5}{n}$ are zero, as ${n\to\infty}$, so the expression just becomes $\frac{2}{-3}$

i.e. $\frac{-2}{3}$.

Hope it helps!

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$A(n) = \frac{2n^2 + 3n}{5-3n^2} $ so

$\begin{array}\\ A(n)+\frac23 &= \frac{2n^2 + 3n}{5-3n^2}+\frac23\\ &= \frac{3(2n^2 + 3n)+2(5-3n^2)}{3(5-3n^2)}\\ &= \frac{6n^2 + 9n+10-6n^2}{3(5-3n^2)}\\ &= \frac{ 9n+10}{15-9n^2}\\ &= \frac{ 9/n+10/n^2}{15/n^2-9}\\ \text{so}\\ |A(n)+\frac23| &= \frac{ 9/n+10/n^2}{|9-15/n^2|}\\ &\le \frac{ 19/n}{|9-15/16|} \qquad\text{for } n \ge 4\\ &\le \frac{ 19}{8n} \qquad\text{for } n \ge 4\\ \end{array} $

The rest should be easy.

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    $\begingroup$ how did -9n^2/n^2 become 3? $\endgroup$ – elsteinm Nov 2 '16 at 20:28
  • $\begingroup$ Thanks. 3x3=3 only in my brain, sometimes. Fixed. Upvoted. $\endgroup$ – marty cohen Nov 2 '16 at 20:32
  • $\begingroup$ also, by the second to last line, 15/n^2 becomes 15/16, is that a mistake? $\endgroup$ – elsteinm Nov 2 '16 at 20:35
  • $\begingroup$ No, because of $n \ge 4$. $\endgroup$ – marty cohen Nov 3 '16 at 14:49

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