I made up a series. oh oh.

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\(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{n^n}{\Gamma(n^2+1)} }\)
should go by `Γ(x+1)=x!` that \(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{n^n}{(n^2)!} }\)
what is r

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it is the gamma function
oh thanks
\(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n+1}}{(n+1)^2!}~\times\frac{n^2!}{n^n} }\) ratio test...
\(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n+1}}{(n+1)^2!}~\times\frac{n^2!}{n^n} }\) \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n+1}}{n^n}~\times\frac{n^2!}{(n+1)^2!} }\) \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{(n+1)^{n}}{n^n}~\times\frac{n^2!~(n+1)}{(n+1)^2!} }\) \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{n^2!~(n+1)}{(n+1)^2!} }\) =0 (by numerical approach.)
therefore series converges
I 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....
nice series man!
\(\normalsize\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\frac{n^2!~(n+1)}{(n+1)^2!} }\) 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 \({\rm k}\in~{\bf R}\))
I compared it to this larger convergent series: \[\sum_{n=1}^\infty \left( \frac{n}{n^2+n-1} \right)^n\] 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.
I 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.)
there'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 $$\sum_{n=N_0}^\infty \frac1{\sqrt{2\pi} n^{2n^2-n+1} e^{-n^2}}=\sqrt{2\pi}\sum_{n=N_0}^\infty\frac{e^{n^2}}{n^{2n^2-n+1}}$$
\(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \frac{e^{n^2}}{n^{2n^2-n+1}} }\) then, comparison test (compariong to the series below which is larger) \(\large\color{black}{ \displaystyle \sqrt{2\pi} \sum_{ n=1 }^{ \infty } \frac{e^{n^2}}{n^{n^2}} }\) re-writing/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 re-writing 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.
2,5,10,?
can you figure out the next number?
17 :P

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