[SOLVED] Find the interval of convergence of the series

**lilaziz1** · April 7th 2010, 12:57 PM

55. $\sum_{n=1}^{\infty} \frac{n^n(x+2)^n}{3^nn!}$

My Solution

$\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} | = \frac{(n+1)^{n+1}(x+2)^{n+1}}{3^{n+1}(n+1)!} * \frac{3^nn!}{n^n(x+2)^n}$

$= \frac{(n+1)(n+1)^n(x+2)}{3(n+1)n^n} \Longrightarrow \frac{|x+2|}{3} < 1 \Longrightarrow -5 < x < 1$

However, the solution from the book goes like this:

Book Solution

$\lim_{n\rightarrow\infty} \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}(x+2)^{n+1}} * \frac{n^n(x+2)^n}{3^nn!}$

$\lim_{n\rightarrow\infty} \frac{3n^n}{(n+1)^n(x+2)} = \frac{3}{x+2}$

$-\frac{3}{e} -2 < x < \frac{3}{e} -2$ or

$-3.104 < x < -0.896$

What am I doing wrong? I copied the solution number/letter by number/letter.

Does the same thing for #56:

56. $\sum_{n=1}^{\infty} \frac{n!x^n}{5^nn^n}$

My Solution

$\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} | = \frac{(n+1)!x^{n+1}}{(n+1)^{n+1}5^{n+1}} * \frac{5^nn^n}{n!x^n}$

$\lim_{n\rightarrow\infty} \frac{|x|}{5} < 1 \Longrightarrow -5 < x < 5$

Book Solution

$\lim_{n\rightarrow\infty} \frac{(n+1)^{n+1}5^{n+1}}{(n+1)!x^{n+1}} * \frac{n!x^n}{5^nn^n}$

$\lim_{n\rightarrow\infty} \frac{(n+1)^n5}{n^nx} = \frac{5}{x}$

$-5e < x < 5e$ or

$= -13.591 < x < 13.591$

any help is appreciated.

**undefined** · April 7th 2010, 01:12 PM

Originally Posted by lilaziz1

55. $\sum_{n=1}^{\infty} \frac{n^n(x+2)^n}{3^nn!}$

My Solution

$\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} | = \frac{(n+1)^{n+1}(x+2)^{n+1}}{3^{n+1}(n+1)!} * \frac{3!n!}{n^n(x+2)^n}$

$= \frac{(n+1)(n+1)^n(x+2)}{3(n+1)n^n} \Longrightarrow \frac{|x+2|}{3} < 1 \Longrightarrow -5 < x < 1$

However, the solution from the book goes like this:

Book Solution

$\lim_{n\rightarrow\infty} \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}(x+2)^{n+1}} * \frac{n^n(x+2)^n}{3^nn!}$

$\lim_{n\rightarrow\infty} \frac{3n^n}{(n+1)^n(x+2)} = \frac{3}{x+2}$

$-\frac{3}{e} -2 < x < \frac{3}{e} -2$ or

$-3.104 < x < -0.896$

What am I doing wrong? I copied the solution number/letter by number/letter.

You used $\lim_{n\rightarrow\infty} |\frac{a_{n+1}}{a_n} |$ but it it's supposed to be $\lim_{n\rightarrow\infty} |\frac{a_n}{a_{n+1}} |$ , like what the book did... does that help?

**zzzoak** · April 7th 2010, 01:19 PM

Please notice, that
$\lim\frac{{(n+1)}^n}{n^n}$ = $\lim{(1+\frac{1}{n})}^n$ = e

**lilaziz1** · April 7th 2010, 01:22 PM

but isn't the Ratio test $\frac{|a_{n+1}|}{|a_n|}$

**lilaziz1** · April 7th 2010, 01:26 PM

:O! i just re-did the problem. thanks a ton zzzoak!

**zzzoak** · April 7th 2010, 01:29 PM

$<br /> \lim\frac{(n+1)^n}{n^n}\frac{(x+2)}{3} \Longrightarrow \frac{e|x+2|}{3} < 1<br />$

**lilaziz1** · April 7th 2010, 01:33 PM

Originally Posted by zzzoak

$<br /> \lim\frac{(n+1)^n}{n^n}\frac{(x+2)}{3} \Longrightarrow \frac{e|x+2|}{3} < 1<br />$

Yea. Didn't use to pay attention to the n's much til now. thanks once again.

**undefined** · April 7th 2010, 02:16 PM

Originally Posted by lilaziz1

but isn't the Ratio test $\frac{|a_{n+1}|}{|a_n|}$

Hmm, you are right, I misread this Wikipedia article and saw that the book started with $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ and made a wrong conclusion. Sorry about that.

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