calculate the radius of convergence and the interval of convergence (with examination of the endpoints)
$$\sum_{n=1}^{\infty}\frac{3^{n+1}}{n}.x^{n}$$

a try...

$$|\frac{a_{n+1}}{a_{n}}|=$$
$$
\Large \frac{\frac{3^{(n+1)+1}}{n+1}}{\frac{3^{n+1}}{n}}=$$
$$\frac{3^{n+2}}{n+1}.\frac{n}{3^{n+1}}=\frac{3}{1}=3$$
is it correct and what i need to do additional?

Question

calculate the radius of convergence and the interval of convergence (with examination of the endpoints)
$$\sum_{n=1}^{\infty}\frac{3^{n+1}}{n}.x^{n}$$

a try...

$$|\frac{a_{n+1}}{a_{n}}|=$$
$$
\Large \frac{\frac{3^{(n+1)+1}}{n+1}}{\frac{3^{n+1}}{n}}=$$
$$\frac{3^{n+2}}{n+1}.\frac{n}{3^{n+1}}=\frac{3}{1}=3$$
is it correct and what i need to do additional?

anonymous · Answer

good work. you are done

anonymous · Answer

well actually you do have one more step
make sure that $|x|<\frac{1}{3}$

anonymous · Answer

i mean you are done, since you get 3 radius of convergence is $\frac{1}{3}$

anonymous · Answer

now replace $x$ by $\frac{1}{3}$ and see what you get for the sum. 
and repeat with $x=-\frac{1}{3}$

anonymous · Answer

I heard there is a definition for that to find its like Limit / 1  but i didnt see, can you post for me a link where definition is there ?

anonymous · Answer

gotta run, last part should be straightforward, since you are just being asked if a series converges

anonymous · Answer

ok satellite thank you,