Series

**xfyz** · November 4th 2007, 03:59 PM

Find the radius of convergence and interval of convergence of the series:

Sum of n(x-4)^n / (n^3+1) with n = 1 to infinity.

can someone help me with this? Thanks!

**Plato** · November 4th 2007, 05:16 PM

$a_n = \frac{n}{{n^3 + 1}}\left( {x - 4} \right)^n$

$\left( {\sqrt[n]{{\left| {a_n } \right|}}} \right) = \left( {\sqrt[n]{{\frac{n}{{n^3 + 1}}}}\left| {x - 4} \right|} \right) \to \left| {x - 4} \right|<br />$

**Soroban** · November 4th 2007, 05:58 PM

Hello, xfyz!

Find the radius of convergence and interval of convergence of the series:

. . $\sum^{\infty}_{n=1}\frac{n(x -4)^n}{(n^3+1)}$

I'd use the Ratio Test . . .

$\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{(n+1)(x-4)^{n+1}}{(n+1)^3+1} \cdot \frac{n^3+1}{n(x-4)^n}\right| \;=$ . $\left|\frac{n+1}{n}\cdot\frac{n^3+1}{n^3+3n^2+3n + 2}\cdot\frac{(x-4)^{n+1}}{(x+4)^n}\right|$

Divide top and bottom of the second fraction by $n^3$

. . $\left|\left(1 + \frac{1}{n}\right)\left(\frac{1 + \frac{1}{n^3}}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{2}{n^3}}\right)(x-4)\right|$

Take the limit:

. . $\lim_{n\to\infty} \left|\left(1 + \frac{1}{n}\right)\left(\frac{1 + \frac{1}{n^3}}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{2}{n^3}}\right)(x-4)\right| \;=\;\left|(1 + 0)\left(\frac{1+0}{1+0+0+0}\right)(x-4)\right|$ . $\;=\;|x-4|<br />$

We have: . $|x - 4| \:< \:1\quad\Rightarrow\quad -1 \:<\:x-4\:<\:1\quad\Rightarrow\quad 3 \:<\:x\:<\:5$

. . The radius of convergence is: . $R \,=\,1$

. . The interval of convergence: . $(3,\,5)$

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