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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! :))

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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
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@MichaelHardy: Certainly. But the substitution may bring the student into more familiar territory. –  André Nicolas Dec 25 '12 at 19:46
    
As @DonAntonio says, 10 questions, 0 accepted answers. You would do well to upvote helpful answers and accept one for all previous questions with acceptable answers. –  JohnD Dec 25 '12 at 20:33
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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 $x$.

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