Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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! :))

share|improve this question
(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

1 Answer 1

up vote 1 down vote accepted

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 $x$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.