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SolomonZelman
 one year ago
I made up a series. oh oh.
SolomonZelman
 one year ago
I made up a series. oh oh.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11\(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{n^n}{\Gamma(n^2+1)} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11should go by `Γ(x+1)=x!` that \(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{n^n}{(n^2)!} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11it is the gamma function

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11\(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n+1}}{(n+1)^2!}~\times\frac{n^2!}{n^n} }\) ratio test...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11\(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n+1}}{(n+1)^2!}~\times\frac{n^2!}{n^n} }\) \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n+1}}{n^n}~\times\frac{n^2!}{(n+1)^2!} }\) \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n}}{n^n}~\times\frac{n^2!~(n+1)}{(n+1)^2!} }\) \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{n^2!~(n+1)}{(n+1)^2!} }\) =0 (by numerical approach.)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11therefore series converges

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11I will continue to tackle this prob next week. If you want to look at a mentally disabled always welcome:) I am going offline right now....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11\(\normalsize\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{n^2!~(n+1)}{(n+1)^2!} }\) how does it behave? \(\normalsize\color{black}{ \displaystyle \frac{\color{blue}{1}^2!~(\color{blue}{1}+1)}{(\color{blue}{1}+1)^2!}=\frac{1!(2)}{4!} =1/12 }\) \(\normalsize\color{black}{ \displaystyle \frac{\color{blue}{2}^2!~(\color{blue}{2}+1)}{(\color{blue}{2}+1)^2!}=\frac{4!(3)}{9!} =1/(5\cdot6\cdot7\cdot8\cdot3) }\) \(\normalsize \color{black}{ \displaystyle \frac{\color{blue}{3}^2!~(\color{blue}{3}+1)}{(\color{blue}{3}+1)^2!}=\frac{9!(4)}{16!} =1/(10\cdot11\cdot12\cdot13\cdot14\cdot15\cdot4) }\) \(\normalsize \color{black}{ \displaystyle \frac{\color{blue}{4}^2!~(\color{blue}{4}+1)}{(\color{blue}{4}+1)^2!}=\frac{16!(5)}{25!} =1/(17\times ~~{\bf ....}~~\times24\times5) }\) \(\normalsize \color{black}{ \displaystyle \frac{\color{blue}{5}^2!~(\color{blue}{5}+1)}{(\color{blue}{5}+1)^2!}=\frac{25!(6)}{36!} =1/(26\times ~~{\bf ....}~~\times 35\times6) }\) the common ratio clearly becomes 0. You are multiplying by the tiniest values you can (and even can't) imagine.So the common ratio is r=0. ... By the way \(\Gamma \) is a Latin latter originally (but it is also Russian). Also, from here you clearly see a way to simplify the limit, which you wouldn't necessarily think of (or understand if showed) without expanding the expression after plugging in values for n the way I showed for n=1, 2, 3, 4, 5.. \(\normalsize\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{n^2!~(n+1)}{(n+1)^2!} ~~~{\LARGE =} \\[1.7em] \displaystyle \lim_{n\rightarrow \infty}\frac{n+1}{(n^2+1) \times (n^2+2)\times ~{\bf ....} \times ~(n+1)^2}}\) I hope it is clear where I am going. n+1 cancel, and then we have a polynomial with n's  in fact a huge polynomial with n's. And that limit is thus 0. Very easy to tell it =0, in fact: \(\normalsize\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{{\rm C} }{f(n)}=0}\)\(\normalsize\color{black}{ ; \\[1.5em]}\) where \(\color{black}{f(n)} \) is a function that increases over (∞,+∞). in fact even if increases over [k,+∞) (k some defined  even very big  constant of any value, and \({\rm k}\in~{\bf R}\))

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I compared it to this larger convergent series: \[\sum_{n=1}^\infty \left( \frac{n}{n^2+n1} \right)^n\] on the reasoning that: \[\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*\frac{1}{6!} < \left( \frac{3}{7} \right)^3\]\[\frac{4}{16}*\frac{4}{15}*\frac{4}{14}*\frac{4}{13}*\frac{1}{12!} < \left( \frac{4}{13} \right)^4\] etc... Well I am not fully explaining but hopefully this gets my point across rather than go into detail.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11I didn't see any comparison tests, \(\rm \color{royalblue}{thanks~for~your~suggesstion}\). This new series is designed for the nth root test (performing this test will allow us to see the common ratio as n→∞. (The limit set up by the compariosn test is equal to zero, showing divergence.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there's probably some value investigating it using the Stirling formula \(n!\sim\sqrt{2\pi n}\ n^ne^{n}\) we have that \((n^2)!\sim \sqrt{2\pi} n^{2n^2+1} e^{n^2}\), so our sum behaves in the long run like $$\sum_{n=N_0}^\infty \frac1{\sqrt{2\pi} n^{2n^2n+1} e^{n^2}}=\sqrt{2\pi}\sum_{n=N_0}^\infty\frac{e^{n^2}}{n^{2n^2n+1}}$$

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.11\(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \frac{e^{n^2}}{n^{2n^2n+1}} }\) then, comparison test (compariong to the series below which is larger) \(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \frac{e^{n^2}}{n^{n^2}} }\) rewriting/simplifying: \(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \left(\frac{e^n}{n^n}\right)^n }\) this way, there is a common ratio of e^n/n^n Or, without rewriting it this way, \(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \left(\frac{e}{n}\right)^{n^2}=a_1+a_4+a_9+a_{16}+... }\) \(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \left(\frac{e}{n}\right)^{n^2}=e+\frac{e^4}{2^4}+\frac{e^9}{3^9}+\frac{e^{16}}{4^{16}}+\frac{e^{25}}{5^{25}}+... }\) it is clearly convergent.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0can you figure out the next number?
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