# Divergent series exercises

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?

• The first converges for small $a$ does it not? It has terms smaller than a geometric series. – oshill Nov 21 '19 at 23:47

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

$$\frac{a^n}{b^n+1} < a^n$$