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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?

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

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

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