1. ## Summation help (urgent)

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

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

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

$\displaystyle \sum_{r=1}^{\infty} \frac{1}{3^{r-2} r}$
What are you trying to do with it? Show it converges or sum it?

3. 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?

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

5. Originally Posted by Prove It
Well obviously if you evaluate the sum, it's convergent... So yes, you'd show both.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{3^{n-2}n}\leq\sum_{n=1}^{\infty}\frac{1}{3^{n-2}}=\frac{9}{2}$

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

$\displaystyle 9\ln\left(\frac{3}{2}\right)$

6. Originally Posted by intomath
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?
Finding the value of the sum will answer the question, which is all that you're required to do.

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

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

8. Originally Posted by mr 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 $\displaystyle |r| < 1$.
$\displaystyle -\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}~|x|<1\cup\left \{-1\right\}$

...just saying