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?

  • $\begingroup$ 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 $1\cdot 0=0$. $\endgroup$ – Paramanand Singh Sep 28 '18 at 5:37

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.

  • 1
    $\begingroup$ Hi Olivier ! Long time no see. Cheers. $\endgroup$ – Claude Leibovici Sep 28 '18 at 5:22
  • $\begingroup$ Claude, je vous souhaite un excellent week-end ! $\endgroup$ – 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 $n>6$.


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