Determine the result of


I would like to use the sandwich rule for limits to find two sequences which define lower and upper bounds to determine the appropriate limit which is obviously $+\infty$.

By now I have an upper bound:

$$\frac{n^{n+1}}{n!}=n\cdot\frac{n}{n}\cdot\frac{n}{n-1}\cdot\ldots\cdot\frac{n}{2}\cdot\frac{n}{1}\leq n\cdot 1\cdot n\cdot\ldots\cdot n\cdot n=n^n\longrightarrow+\infty$$

What would you suggest, which sequence should I use for a lower bound which converges to $+\infty$?


As your sequence is positive it suffices to find a lower bound sequence that diverges to infinity:

$$\frac{n^{n+1}}{n!}=n\frac{n\cdot n\cdot\ldots\cdot n}{1\cdot 2\cdot\ldots\cdot n}\geq n\frac{n\cdot n\cdot\ldots\cdot n}{n\cdot n\cdot\ldots\cdot n}=n\xrightarrow [n\to\infty]{} \infty$$

  • 1
    $\begingroup$ What would you do if it was (n^n)/n! $\endgroup$ – Adam Rubinson Nov 5 '12 at 13:56
  • $\begingroup$ I'd take out $\,\dfrac{n}{1}\,$ from the fraction: $$\frac{n^n}{n!}\geq \frac{n}{1}\frac{n\cdot...\cdot n}{n\cdot\ldots\cdot n}=n....$$ $\endgroup$ – DonAntonio Nov 5 '12 at 14:05
  • $\begingroup$ lol! And if it were (n)^(n-k) / n! (for some fixed k) ? $\endgroup$ – Adam Rubinson Nov 5 '12 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.