Series
\[\sum_{n=1}^{\infty} \frac{ n! }{ n^n }\]
@ganeshie8
I did ratio test and got 1, but isn't that inconclusive then, but if I use ratio test and harmonic series I got divergent. But it says with ratio test it's convergent, mhm?
ratio test is giving me the limit 1/e
1/e < 1 so the series converges
check your work again
ill brb in 5 min and ill show my steps, sorry!
k
\[\huge \lim_{n \rightarrow \infty} \left| \frac{ \frac{ (n+1)! }{ (n+1)^{n+1} } }{ \frac{ n! }{ n^n } } \right|\] \[\huge \lim_{n \rightarrow \infty} \left| \frac{ (n+1)(n)! }{ (n+1)^n(n+1) } \times \frac{ n^n }{ n! } \right|\]
\[\huge \lim_{n \rightarrow \infty} \left| \frac{ n^n }{ (n+1)^n } \right| \implies \lim_{n \rightarrow \infty} \left( \frac{ n }{ (n+1) } \right)^n\]
\[\huge \lim_{n \rightarrow \infty} \left( \frac{ n/n }{ n/n+1/n } \right)^{n/n} \implies 1\] this is what I did or am I suppose to do the y = (x/(x+1))^x business?
how on earth are you allowed to divide the exponent by n ?
Lol I had a feeling I wasn't suppose to do that
So it's an indeterminate form right?
I can see where this is going now and how to get 1/e
Don't judge me! I know that was a rookie mistake, I was being lazy!
if you want an indeterminate form : \[ \large \lim_{n \rightarrow \infty} \left( \frac{ n }{ n+1 } \right)^n = e^{\lim\limits_{n \rightarrow \infty} n \ln \left( \frac{ n }{ n+1} \right) }\]
Yeah haha I see that
however you can just use the standard limit (1+1/n)^n = e
I was trying to slip through without doing much work
Yeah every time I see something like that I think of e, I knew it had to do something with it, but I instead embarrassed myself
let wolfram do the limits
sup
Alright lol, thanks :D Now lets not tell anyone about this.
This one is pretty tricky if you flip the question, you'll get e and then divergent, I guess I was thinking about that instead
the site is acting up
i know i can text or receive texts
Yeah, I keep losing connection and what not. They need to really fix the site, or reboot if that's possible haha.
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