can someone help me go through this question i cant figure it out, i need help asap :(

$\displaystyle \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r} $

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- Nov 11th 2008, 11:11 AMintomathSummation help (urgent)
can someone help me go through this question i cant figure it out, i need help asap :(

$\displaystyle \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r} $

- Nov 11th 2008, 12:35 PMMathstud28
- Nov 11th 2008, 02:17 PMintomath
all it says in my paper is Find

$\displaystyle

\sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}

$

It carries 8 marks though, maybe you have to show both? - Nov 11th 2008, 02:39 PMProve It
- Nov 11th 2008, 02:54 PMMathstud28
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{3^{n-2}n}\leq\sum_{n=1}^{\infty}\frac{1}{3^{n-2}}=\frac{9}{2}$

Now for the sum rewrite it as

$\displaystyle 9\sum_{n=1}^{\infty}\frac{\left(\frac{1}{3}\right) ^n}{n}$

Now it is known that

$\displaystyle \sum_{n=1}^{\infty}\frac{\left(x\in[-1,1)\right)^n}{n}=-\ln(1-x)$

So our sum is

$\displaystyle 9\ln\left(\frac{3}{2}\right)$ - Nov 11th 2008, 03:32 PMmr fantastic
- Nov 11th 2008, 03:55 PMmr fantastic
Sum of infinite geometric series:

$\displaystyle \sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ for $\displaystyle |r| < 1$.

Integrate both sides (there are some technical issues regarding the justification of changing the order of operations):

$\displaystyle \sum_{n=0}^{\infty} \frac{r^{n+1}}{n+1} = - \ln (1 - r) + C$ for $\displaystyle |r| < 1$

Substitute $\displaystyle r = 0: ~ 0 = 0 + C \Rightarrow C = 0$

$\displaystyle \Rightarrow \sum_{n={\color{red}1}}^{\infty} \frac{r^n}{n} = - \ln (1 - r)$ for $\displaystyle |r| < 1$. - Nov 11th 2008, 06:16 PMMathstud28