1. ## Series

$\displaystyle \left(\frac{1}{1}\right)$+$\displaystyle \left(\frac{1+2}{1.2}\right)$+$\displaystyle \left(\frac{1+2+3}{1.2.3}\right)$+$\displaystyle \left(\frac{1+2+3+4}{1.2.3.4}\right)$+...................
find out the $\displaystyle \bigcup r$. and put it into partial fraction. Through this $\displaystyle \sum_{r=1}^{\propto} \bigcup r$=$\displaystyle \frac{3e}{2}$ In this case i write the $\displaystyle \bigcup{r}$
like this $\displaystyle \bigcup r$=$\displaystyle \frac{\sum_{r=1}^{\propto}}{r!}$ Then how can i partial fraction.>>>>>>

2. ## Re: Series

A little correction $\displaystyle \bigcup{r}=\frac{\sum_{r=1}^{\propto}r}{r!}$

3. ## Re: Series

I have no clue what you are asking. Please try stating what you are trying to prove again? I am not sure where partial fractions comes into this discussion. Nor do I have any idea what the end result is supposed to be.

4. ## Re: Series

I ask my $\displaystyle \bigcup{r}=\frac{\sum_{r=1}^{\propto}r}{r!}$ Is correct or not, and then How can i put it into partial fractions.......

5. ## Re: Series

$\displaystyle \bigcup r$ means the union of the sets $\displaystyle r$. $\displaystyle \sum_{r=1}^\infty r$ is not defined on sets. Then, you are dividing by $\displaystyle r!$. The notation you are using is meaningless. The only place you define $\displaystyle r$ is in the sum. But then you use $\displaystyle r$ outside the sum, as well. Nowhere do you have anything that would allow partial fractions to be applicable.

Edit: from the series you tried to write, it looks like you are looking for:

$\displaystyle \sum_{n\ge 1}\dfrac{\sum_{r = 1}^n r}{n!}$

If that is what you mean, then what do you think partial fractions are?

6. ## Re: Series

Here is something you can do, but it is not partial fractions:

\displaystyle \begin{align*}& 1 + \dfrac{1+2}{2!} + \dfrac{1+2+3}{3!} + \cdots \\ = & \sum_{n\ge 1}\dfrac{\binom{n+1}{2}}{n!} \\ = & \sum_{n\ge 1}\dfrac{\left(\dfrac{(n+1)!}{2!(n-1)!}\right)}{n!} \\ = & \sum_{n\ge 1}\dfrac{(n+1)}{2!(n-1)!} \\ = & \sum_{n\ge 1}\dfrac{(n-1)+2}{2!(n-1)!} \\ = & \sum_{n\ge 1}\dfrac{n-1}{2!(n-1)!} + \sum_{n\ge 1}\dfrac{2}{2!(n-1)!} \\ = & \dfrac{1}{2}\sum_{n\ge 0}\dfrac{n}{n!} + \sum_{n\ge 0}\dfrac{1}{n!} \\ = & \dfrac{1}{2}\sum_{n\ge 0}\dfrac{1}{n!} + e \\ = & \dfrac{e}{2} + e \end{align*}

That gets you what you want.