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How can I check convergence of $ \sum_{n=1}^{\infty} 2^n \cdot \left(\frac{n}{n+1}\right)^{n^2} $ ? If I want check necessary condition $u_n \rightarrow 0$ I need to do sth like that: $$ u_n = 2^n \cdot \left(\frac{n}{n+1}\right)^{n^2} = 2^n \cdot \left(\left(\frac{1}{1+\frac{1}{n}}\right)^{n}\right)^2 $$ but now I can't write just $$ \left(\left(\frac{1}{1+\frac{1}{n}}\right)^{n}\right)^2 \rightarrow \frac{1}{e^2}$$ because I have $ 2^n $ part... Thanks for your time.

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    $\begingroup$ Hmmm... $$\left(\frac{n}{n+1}\right)^{n^2} \ne \left(\left(\frac{1}{1+\frac{1}{n}}\right)^{n}\right)^2=\left(\frac{n}{n+1}\right)^{2n} $$ $\endgroup$ – Did Dec 13 '18 at 16:27
  • $\begingroup$ Ahh.. My fail, your right, thanks $\endgroup$ – VirtualUser Dec 13 '18 at 16:30
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Using Root test,

$$\lim_{n\to\infty}\sqrt[n]{u_n}=\dfrac2{\lim_{n\to\infty}\left(1+\dfrac1n\right)^n}=\dfrac2e<1$$

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  • $\begingroup$ It is great idea! $\frac{1}{e} \in \left(\frac{1}{2};\frac{1}{3} \right) $ so it is clearly under $1$. But how can I solve necessary condition? I can't use there root test $\endgroup$ – VirtualUser Dec 13 '18 at 16:32

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