# convergence $\sum_{n=1}^{\infty} 2^n \cdot \left(\frac{n}{n+1}\right)^{n^2}$

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.

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

$$\lim_{n\to\infty}\sqrt[n]{u_n}=\dfrac2{\lim_{n\to\infty}\left(1+\dfrac1n\right)^n}=\dfrac2e<1$$
• 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 – VirtualUser Dec 13 '18 at 16:32