# Determining the radius of convergence

Oh sorry everyone, I have just written the problem. Thanks for the warning.

$$\sum_{n=1}^{\infty}\frac{(x+3)^n}{n\cdot 3^{n}}$$

I could not determine the radius of convergence and the interval of convergence of the series where n starts at 1 and goes up to infinity.

Can anyone help me? Thanks for any help! :))

-
(1) Enhance seriously your accept rate. People tend not to invest their free time to try to help people who doesn't show appreciation for their efforts; (2) Show some self work, some effort, ideas...(3) Using the word "please" won't hurt. –  DonAntonio Dec 25 '12 at 19:28
As a start, replace $x+3$ by $t$. Then Ratio Test works nicely. Root Test also. –  André Nicolas Dec 25 '12 at 19:42
@AndréNicolas : It's not really necessary to do that substitution. The ratio test works regardless of whether you do that. –  Michael Hardy Dec 25 '12 at 19:44
@MichaelHardy: Certainly. But the substitution may bring the student into more familiar territory. –  André Nicolas Dec 25 '12 at 19:46
oh i see. thank you for the explanantion @johnD –  Yigit Can Dec 25 '12 at 23:46
show 1 more comment

We're looking at $$\sum_{n=0}^\infty \frac{(x+3)^n}{n\cdot 3^n}.$$
Applying the ratio test, we have $$\lim_{n\to\infty} \left| \frac{\left(\frac{(x+3)^{n+1}}{(n+1)\cdot 3^{n+1}}\right)}{\left(\frac{(x+3)^n}{n\cdot 3^n}\right)} \right| = \lim_{n\to\infty} \left|\frac{n(x+3)}{3(n+1)}\right| = \lim_{n\to\infty} \left(\frac{|x+3|}{3}\cdot\frac{n}{n+1}\right).$$ The factor $\dfrac{|x+3|}{3}$ does not change as $n$ changes, so it can be pulled out: $$=\frac{|x+3|}{3} \lim_{n\to\infty} \frac{n}{n+1} = \frac{|x+3|}{3}\cdot 1$$
Thus the series converges if $\dfrac{|x+3|}{3}<1$ and diverges if $\dfrac{|x+3|}{3}>1$.
Now solve the inequality $$\frac{|x+3|}{3}<1$$ for $x$.