radius of convergence: sum x^k/k^k

Question

australopithecus · Answer

are we talking about series here?

australopithecus · Answer

ugh its k^k

anonymous · Answer

(K+1)^k+1 right?

australopithecus · Answer

same deal

$$\lim_{k \rightarrow \infty} |\frac{\frac{x^{k+1}}{(k+1)^{k+1}}}{\frac{x^k}{k^k}}|$$

=

$$\lim_{k \rightarrow \infty} |\frac{x^kxk^k}{x^k(k+1)^k(k+1)}|$$

=

$$\lim_{k \rightarrow \infty} |\frac{xk^k}{k^k(1+\frac{1}{k})^k(k+1)}|$$

=

$$\lim_{k \rightarrow \infty} |\frac{x}{(1+\frac{1}{k})^k(k+1)}|$$

hmmm

australopithecus · Answer

$$|\frac{x}{{1}^{\infty}(\infty+1)}| =  |\frac{x}{{1}(\infty)}| =  |\frac{x}{\infty}|$$

australopithecus · Answer

Here is a video on the subject https://www.youtube.com/watch?v=MM3BXtVu9eM I hope this was helpful

australopithecus · Answer

any questions?

anonymous · Answer

no questions, thanks. this is very helpful

australopithecus · Answer

Wait no I was right to begin with I think

australopithecus · Answer

I made a mistake in wolfram alpha the radius is in fact infinity

australopithecus · Answer

https://www.wolframalpha.com/input/?i=series+k+%3D+1+to+infinity+x^k%2Fk^k you can type any value for x and it will converge $$|\frac{x}{\infty}| < 1$$ $$|{x}| < \infty$$ therefore the radius is infinity

australopithecus · Answer

sorry about the confusion but yeah that is the answer

anonymous · Answer

good catch, thanks

australopithecus · Answer

no problem thanks for the question