3
$\begingroup$

Immediately I recognize that there's a factorial and I use the ratio test to try and solve it:

$$\lim_{n \rightarrow \infty}\left|\frac{{(n+1)}^{\sqrt{n+1}}}{(n+1)!}\cdot\frac{n!}{n^{\sqrt{n}}}\right|=\lim_{n \rightarrow \infty}\left|\frac{{(n+1)}^{\sqrt{n+1}-1}}{n^{\sqrt{n}}}\right|$$

At this point I'm not sure how to evaluate the limit. The answer key says the limit is 0, but how is it getting that? Is there an easier way to see that the series converges?

$\endgroup$
  • $\begingroup$ Try to show that $(n+1)^{\sqrt{n+1}} \sim n^{\sqrt{n}}$ (the limit of the ratio of these two terms goes to $1$) then the limit simplifies to $\frac{1}{n^{n-\sqrt{n}}}$ for which it's easy. $\endgroup$ – Winther Jun 22 '16 at 17:05
  • $\begingroup$ It may be easier to show that the series converges using Sterling's approximation $\endgroup$ – Zubin Mukerjee Jun 22 '16 at 17:05
  • $\begingroup$ @ZubinMukerjee It is "Stirling's," not "Sterling's." $\endgroup$ – Mark Viola Jun 22 '16 at 17:11
  • $\begingroup$ @Dr.MV Thanks ... too late to edit the comment, unfortunately :( $\endgroup$ – Zubin Mukerjee Jun 22 '16 at 17:13
  • $\begingroup$ You're missing a square root in your expression : $\frac{(n+1)^{\sqrt{n+1}}}{(n+1)!}\cdot \frac{n!}{n^{\sqrt n}}$ (second denominator). $\endgroup$ – Christoph Frings Jun 22 '16 at 17:21
2
$\begingroup$

$\displaystyle\sum_{n=1}^{\infty}\frac{2^n}{n!}$ converges by the Ratio Test $\;\;$ (since $\displaystyle\lim_{n\to\infty}\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}=\lim_{n\to\infty}\frac{2}{n+1}=0<1)$;

and $\displaystyle\ln n\le \sqrt{n}\ln 2\implies \sqrt{n}\ln n\le n\ln 2\implies n^{\sqrt{n}}\le 2^n\implies\frac{n^{\sqrt{n}}}{n!}\le\frac{2^n}{n!}$ for $n$ large,

so $\displaystyle\sum_{n=1}^{\infty}\frac{n^{\sqrt{n}}}{n!}$ converges by the Comparison Test.

$\endgroup$
1
$\begingroup$

$n!\geq \left(\frac{n}{e}\right)^n$ is simple to prove, and it trivially gives that the series is convergent.

$\endgroup$
0
$\begingroup$

Note that we can write

$$\begin{align} \frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}&=e^{(\sqrt{n+1}-\sqrt{n})\log(n)+\sqrt{n+1}(\log(n+1)-\log(n))}\\\\ &=e^{\left(\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\right)}e^{\sqrt{n+1}\log\left(1+\frac1n\right)}\\\\ &=e^{\left(\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\right)}e^{\sqrt{n+1}\,O\left(\frac1n\right)}\\\\ &\to 1\,\,\text{as}\,\,n\to \infty \end{align}$$

$\endgroup$
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Jun 22 '16 at 17:45
0
$\begingroup$

You can evuluate limit as follows: $$\lim _{ n\rightarrow \infty } \left| \frac { { (n+1) }^{ \sqrt { n+1 } } }{ (n+1)! } \cdot \frac { n! }{ n^{ n } } \right| =\lim _{ n\rightarrow \infty } \left| \frac { { (n+1) }^{ \sqrt { n+1 } -1 } }{ n^{ n } } \right| =\lim _{ n\rightarrow \infty } \left| \frac { { \left( n+1 \right) }^{ \frac { n }{ \sqrt { n+1 } +1 } } }{ { n }^{ n } } \right| =\\ =\lim _{ n\rightarrow \infty } \left| \left[ { { \left( 1+\frac { 1 }{ n } \right) }^{ n } } \right] ^{ \frac { 1 }{ \sqrt { n+1 } +1 } } \right| =\lim _{ n\rightarrow \infty }{ e } ^{ \frac { 1 }{ \sqrt { n+1 } +1 } }=1$$ but we know in case $1$ test doens't say about converges or diverges

$\endgroup$
  • $\begingroup$ This development has a flaw. $\endgroup$ – Mark Viola Jun 22 '16 at 17:21
0
$\begingroup$

Consider $$A_n=\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}$$ Take logarithms $$\log(A_n)={\sqrt{n+1}}\log(n+1)-{\sqrt{n}}\log(n)$$ $$\log(A_n)={\sqrt{n}} \sqrt{1+\frac 1 n}\left(\log(n)+\log(1+\frac 1 n)\right)-{\sqrt{n}}\log(n)$$ Now, let us use for large values of $n$ $$\sqrt{1+\frac 1 n}=1+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ $$\log(1+\frac 1 n)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and replace. This gives $$\log(A_n)=\frac{1+\frac{1}{2}\log (n)}{\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right)$$ So, when $n\to \infty$, $\log(A_n)\to 0$ and $A_n\to 1$.

$\endgroup$

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.