studying the convergence of the series $\sum_{n=1}^\infty \frac {3^n}{n^32^n}$

Question

I'm studying the convergence of the series $$\sum_{n=1}^\infty \frac {3^n}{n^32^n}$$

• $\frac {3^n}{n^32^n}>0, \forall n \ge 1$
• $\lim_{n \to +\infty}\frac {3^n}{n^32^n}$ should be $0$ .

In another post I've seen that $\frac {3^n}{n^3}>2^n \rightarrow \frac {3^n}{n^32^n}>1 \rightarrow \lim_{n\rightarrow \infty}\frac {3^n}{n^32^n} \ne 0$ and the given serie diverges. But I could proof that $\frac {3^n}{n^3}>2^n $ ?

Answer 1

Considering Cauchy root test we obtain: $$\lim_{n\to\infty}|a_n|^{1\over n}=\lim_{n\to\infty}\frac{1.5}{(n)^\frac{3}{n}}=1.5>1$$ which implies on divergence of the corresponding series.

Answer 2

By ratio test we get $$\frac { { a }_{ n+1 } }{ { a }_{ n } } =\frac { \frac { { 3 }^{ n+1 } }{ { \left( n+1 \right) }^{ 3 }{ 2 }^{ n+1 } } }{ \frac { { 3 }^{ n } }{ { n }^{ 3 }{ 2 }^{ n } } } =\frac { 3 }{ 2 } { \left( 1-\frac { 1 }{ n+1 } \right) }^{ 3 }\quad \overset { n\rightarrow \infty }{ \longrightarrow } \frac { 3 }{ 2 } >1$$

Answer 3

Define $$a_n=\frac{3^n}{n^32^n}$$

If $n\ge 10$, $$\frac{(n+1)^3}{n^3}=\left(1+\frac1{n}\right)^3\le1.1^3<\frac32$$

Since $$\frac{a_{n+1}}{a_n}=\frac32\cdot\frac{n^3}{(n+1)^3}$$ the sequence $\{a_n\}$ is increasing for $n\ge 10$. Since its terms are positive, their sum will hardly converge.

Answer 4

I found a solution applying the induction to $\frac {3^n}{n^3}>2^n$

• $P(1)$ is evidently true
• let's suppose $P(n)$ true
• $2^{n+1}=2^n2<\frac {3^n}{n^3}2<\frac {3^n}{n^3}3<\frac {3^n}{(n+1)^3}3 =\frac {3^{n+1}}{(n+1)^3} \rightarrow P(n+1) $ true