# Summation help (urgent)

• November 11th 2008, 11:11 AM
intomath
Summation help (urgent)
can someone help me go through this question i cant figure it out, i need help asap :(

$\sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}$
• November 11th 2008, 12:35 PM
Mathstud28
Quote:

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

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

What are you trying to do with it? Show it converges or sum it?
• November 11th 2008, 02:17 PM
intomath
all it says in my paper is Find
$

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

It carries 8 marks though, maybe you have to show both?
• November 11th 2008, 02:39 PM
Prove It
Quote:

Originally Posted by intomath
all it says in my paper is Find
$

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

It carries 8 marks though, maybe you have to show both?

Well obviously if you evaluate the sum, it's convergent... So yes, you'd show both.
• November 11th 2008, 02:54 PM
Mathstud28
Quote:

Originally Posted by Prove It
Well obviously if you evaluate the sum, it's convergent... So yes, you'd show both.

$\sum_{n=1}^{\infty}\frac{1}{3^{n-2}n}\leq\sum_{n=1}^{\infty}\frac{1}{3^{n-2}}=\frac{9}{2}$

Now for the sum rewrite it as

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

Now it is known that

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

So our sum is

$9\ln\left(\frac{3}{2}\right)$
• November 11th 2008, 03:32 PM
mr fantastic
Quote:

Originally Posted by intomath
all it says in my paper is Find
$

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

It carries 8 marks though, maybe you have to show both?

Finding the value of the sum will answer the question, which is all that you're required to do.
• November 11th 2008, 03:55 PM
mr fantastic
Quote:

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

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

Sum of infinite geometric series:

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

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

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

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

$\Rightarrow \sum_{n={\color{red}1}}^{\infty} \frac{r^n}{n} = - \ln (1 - r)$ for $|r| < 1$.
• November 11th 2008, 06:16 PM
Mathstud28
Quote:

Originally Posted by mr fantastic
Sum of infinite geometric series:

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

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

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

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

$\Rightarrow \sum_{n={\color{red}1}}^{\infty} \frac{r^n}{n} = - \ln (1 - r)$ for $|r| < 1$.

$-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}~|x|<1\cup\left \{-1\right\}$

...just saying