- #1

Benny

- 584

- 0

[tex]

\sum\limits_{n = 1}^\infty {\frac{{n!}}{{n^n }}}

[/tex]

[tex]

\sum\limits_{n = 1}^\infty {\frac{{n^2 }}{{e^n }}}

[/tex]

For the first one I tried:

[tex]

\left| {\frac{{a_{n + 1} }}{{a_n }}} \right| = \left| {\frac{{\left( {n + 1} \right)!}}{{n^{n + 1} }}\frac{{n^n }}{{n!}}} \right| = \frac{{n + 1}}{n} \to 1

[/tex] as n tends to infinity so wouldn't the ratio test be inconclusive because the limit is 1? The answer tells me that the limit is supposed to be exp(-1) which is less than one so the series converges. The answer for the second one is the same. So I try to manipulate the summations so that they look like the one for the exponential.

[tex]

e^x = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}

[/tex]

[tex]

= \sum\limits_{n = 0}^\infty {\left( {\frac{{n!}}{{x^n }}} \right)^{ - 1} }

[/tex]

[tex]

e^n = \sum\limits_{n = 0}^\infty {\left( {\frac{{n!}}{{n^n }}} \right)^{ - 1} }

[/tex]

But how would I find the limit of (a_n+1)/(a_n) to be exp(-1)? Can someone help me out?