# $\lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$ and …

To find limits:

$$(a) \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$$

$$(b) \lim_{n \to \infty} \frac{n^3}{n!}$$

For the first one the sequence is oscillating so it does not converge.

For the 2nd one I used ratio test: Let $$a_n = \frac{n^3}{n!}$$, then $$\frac{a_{n+1}}{a_n} = \frac{n! \times (n+1)^3}{n^3 \times (n+1)!} = (1+ \frac1n)^3 \times \frac{1}{1+n}$$. Thus,

$$|\frac{a_{n+1}}{a_n}| < 1$$, by ratio test limit is $$0$$.

Is the solutions correct?

• The second one is easier than you think. Note that it can be written as $\dfrac{n^2}{(n-1)(n-2)}\cdot\dfrac{1}{(n-3)!}$ and thus tends to $1\cdot 0=0$. – Paramanand Singh Sep 28 '18 at 5:37

## 3 Answers

If one sets $$\displaystyle a_n=\frac{(-1)^n(3-n)}{(3n-5)}$$, one gets $$\lim_{n \to \infty}a_{2n}=-\frac13\color{\red}{\ne}\frac13=\lim_{n \to \infty}a_{2n+1}$$ thus $$\left\{a_n\right\}_{\infty}$$ does not converge.

Your second point is fine.

• Hi Olivier ! Long time no see. Cheers. – Claude Leibovici Sep 28 '18 at 5:22
• Claude, je vous souhaite un excellent week-end ! – Olivier Oloa Sep 29 '18 at 8:16

For the first one, just because it's oscillating doesn't mean it diverges. For example, let $$A(n) = \frac{(-1)^n}{e^n}$$. The limit as $$A \to \infty = 0$$ because the denominator increases without bound.

1) Let's first figure out if the non-oscillating terms converge. $$S(n) = \frac{3-n}{3n-5}$$. As $$n \to \infty$$, $$S(n)$$ approximates $$\frac{-n}{3n} = -\frac13$$. That means as $$n \to \infty$$, the entire sequence alternates between $$\frac13$$ and $$-\frac13$$. Diverges.

2) Perfect use of the ratio test. I just know automatically that factorial increases faster than exponential, which increases faster than polynomial, which increases faster than logarithmic.

Another way for the second is $$\dfrac{n^3}{n!}<\dfrac{n}{n}\dfrac{n}{n-1}\dfrac{n}{n-2}\dfrac{1}{n-3}<1\dfrac{n}{\frac{n}{2}}\dfrac{n}{\frac{n}{2}}\dfrac{1}{\frac{n}{2}}=\dfrac{8}{n}<\varepsilon$$ for $$n>6$$.