1. converging or diverging?

sumation of n=1 to inifinity ((n-1)/(n+1))^n(n+1)

he keeps getting that series diverges due to the divergence test
I keep getting that the series absolutely converges due to the root test

2. well, applying the root test we have to compute $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{n-1}{n+1} \right)^{n+1}=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1-\frac{1}{n}}{1+\frac{1}{n}} \right)^{n}\left( \frac{n-1}{n+1} \right)=\frac{e^{-1}}{e}=\frac{1}{e^{2}}<1.$

so the series converges, and absolutely, since it's a series with positive terms.

3. $\displaystyle \sum_{n=1}^{\infty}\left(\frac{n-1}{n+1}\right)^{n(n+1)}$

Did you try the ratio test?. By the ratio test, we get $\displaystyle {\rho}=e^{-2}$

Which means it converges, since $\displaystyle {\rho}<1$

$\displaystyle \sum_{n=1}^{\infty}\left(\frac{n-1}{n+1}\right)^{n(n+1)}=\frac{1351}{814589}$

4. thank you! that helped in solving a debate we were having.

5. A more highfalutan to do it is to notice that $\displaystyle 0<\left(\frac{n-1}{n+1}\right)^{n(n+1)}<\left(\frac{n}{n+1}\right) ^{n^2}$, and with a little algebraic manipulation, the RHS is equal to $\displaystyle e^{n^2\ln\left(1-\frac{1}{n+1}\right)}$. Notice though that $\displaystyle \ln\left(1-\frac{1}{n+1}\right)\stackrel{n\to\infty}\sim \frac{-1}{n+1}\stackrel{n\to\infty}{\sim}\frac{-1}{n}$ which of course implies that $\displaystyle e^{n^2\ln\left(1-\frac{1}{n+1}\right)}\stackrel{n\to\infty}{\sim}e^ {-n}$. So then we can see that $\displaystyle \sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^2 }$ converges by the limit comparison test with $\displaystyle \sum_{n=1}^{\infty}e^{-n}$, which then means that the actual series converes by the comparison test.