I would be really thankful if someone helped me with these two sequence limits. $$\lim_{n\to\infty} \frac{1}{n^2} \left(2+ \frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}\right)$$ I've tried bounding the second term, but I don't know how to solve it. I think it's zero but I'm not sure about separating the limits.. $$\lim_{n\to\infty} \frac{\sqrt{(n-1)!}}{(1+\sqrt{1}) (1+\sqrt{2})\ldots(1+\sqrt{n})} $$ I've tried using the Stirling formula with this limit, but I'm not sure about how to solve what I get…. Thank you!

  • $\begingroup$ You should post each of these limits in a separate question. (With regard to the second: you should state explicitly "what [you] get" so that people can help you interpret it and won't waste time duplicating your effort.) $\endgroup$ – Blue Jun 2 at 16:30

By the Stolz-Cesro Theorem , one has \begin{eqnarray} &&\lim_{n\to\infty} \frac{1}{n^2} \left(2+ \frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}\right)\\ &=&\lim_{n\to\infty} \frac{\sum_{k=1}^n\frac{(k+1)^k}{k^{k-1}}}{n^2}\\ &=&\lim_{n\to\infty} \frac{\sum_{k=1}^{n+1}\frac{(k+1)^k}{k^{k-1}}-\sum_{k=1}^n\frac{(k+1)^k}{k^{k-1}}}{(n+1)^{2}-n^2}\\ &=&\lim_{n\to\infty} \frac{\frac{(n+2)^{n+1}}{(n+1)^{n}}}{2n+1}\\ &=&\lim_{n\to\infty} \frac{n+2}{2n+1}\frac{(n+2)^{n}}{(n+1)^{n}}\\ &=&\lim_{n\to\infty} \frac{n+2}{2n+1}\left(1+\frac{1}{n+1}\right)^n\\ &=&\frac e2. \end{eqnarray}


For the first $$\frac {(n+1)^n}{n^{n-1}}=(n+1)\left(1+\frac 1n\right)^{n-1}=\frac{n+1}{1+\frac 1n}\left(1+\frac 1n\right)^{n}\to (n+1)e$$

For the second, if you erase all the $+1$s you decrease the denominator and increase the fraction. But it makes the fraction $$\frac{\sqrt{(n-1)!}}{\sqrt{n!}}=\frac 1{\sqrt n} \to 0$$

  • $\begingroup$ Sorry but I don't know how to use that equivalence... Should I replace each term of the sum with it and then calculate the limit? And with the second one... How can I match them if they are different factorials? I'm not sure about it… Thank you $\endgroup$ – Ane Jun 2 at 15:44
  • $\begingroup$ For the first, all the big terms will work like this and the small ones won't matter. It takes some justification but you can replace each term this way. Now you have to sum the series, which will wind up with a factor $n^2$ to cancel the division out front. For the second, when you separate the numerator there aren't any factorials any more. The first term is $\sqrt 1,$ the second $\sqrt 2$, and so on. Match them in order with the terms in the denominator. $\endgroup$ – Ross Millikan Jun 2 at 16:00
  • $\begingroup$ For the first, it depends on how quickly the limit is approached. $\endgroup$ – marty cohen Jun 3 at 1:19

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