Using the necessary condition for convergence,show that these series are divergent $$\sum_{n=1}^\infty \frac{a^n}{b^n+1}=?,a>b>0$$ $$\sum_{n=1}^\infty (\frac{3n}{3n+1})^{n}=?$$
For the second I remember there was a trick with +1 and -1 so that you can show the euler constant e. The answer to the second exercise is $\frac{1}{\sqrt[3]{e}}$. What are the steps to that answer?
For the second one we have that
$$\left(\frac{3n}{3n+1}\right)^{n}=\left[\left(1-\frac{1}{3n+1}\right)^{3n+1}\right]^\frac{n}{3n+1} \to \frac1{e^\frac13}$$
then the series diverge.
For the first one for $a>b>1$
$$\frac{a^n}{b^n+1} \sim \left(\frac a b\right)^n \to \infty$$
for $a>b=1$
$$\frac{a^n}{b^n+1} \sim \frac12a^n \to \infty$$
for $a>1>b$
$$\frac{a^n}{b^n+1} \sim a^n \to 1 \infty$$
for $a=1>b$
$$\frac{a^n}{b^n+1} =\frac{1}{b^n+1}\to 1 $$
the series diverges but for $1>a>b>0$ the series converges by geometric series since
$$\frac{a^n}{b^n+1} < a^n$$