# Sum of n terms of the series $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot 3 \cdot 5 \cdot 7}+\cdots$

I need to find the sum of n terms of the series

$$\frac{1}{1\cdot3}+\frac{2}{1\cdot 3\cdot 5}+\frac{3}{1\cdot 3\cdot 5\cdot 7}+\cdots$$

And I've no idea how to move on. It doesn't look like an arithmetic progression or a geometric progression. As far as I can tell it's not telescoping. What do I do?

• What do the dots in $1.3.5$ stand for? Commented Dec 15, 2015 at 14:36
• You can write the terms as $n\frac{2^{n+1}(n+1)!}{(2n+2)!}$. Not sure how that helps, however. Commented Dec 15, 2015 at 14:37
• I believe they stand for multiplication. Commented Dec 15, 2015 at 14:37
• It's standard in some countries and locals for multiplication @Stef Commented Dec 15, 2015 at 14:38
• By the way, what is an $AP$ or $GP.V_n$ method? Commented Dec 15, 2015 at 14:41

It is telescoping. Consider that: $$\frac{1}{1\cdot 3} = \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right),\quad \frac{2}{1\cdot 3\cdot 5} = \frac{1}{2}\left(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}\right),$$ $$\frac{3}{1\cdot 3\cdot 5\cdot 7}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5}-\frac{1}{3\cdot 5\cdot 7}\right),\quad \ldots$$ so: $$\sum_{k=1}^{n}\frac{k}{(2k+1)!!} = \frac{1}{2}\left(1-\frac{1}{(2n+1)!!}\right).$$ As usual, $(2k+1)!!$ stands for $1\cdot 3\cdot 5\cdot\ldots\cdot (2k+1)$.

• Would make it slightly easier to comprehend the telescope if you kept the $1$ on the right side: $\frac{1}{1}-\frac{1}{1\cdot 3}$, $\frac{1}{1\cdot 3}-\frac{1}{1\cdot3 \cdot 5}$, etc. Also, no clear the OP will follow the double-factorial notation. Commented Dec 15, 2015 at 14:49
• @Jack D'Aurizio strikes again :-)!Seeing you answer after quite some....and as usual you rocked it :-D!Thanks!
– user220382
Commented Dec 15, 2015 at 14:50
• Bravo on this one Jack. I was deriving a series involving Erf and some exponential which, as you would have it, reduces to $1/2$ after plugging in a value of $x$. But this is way way more elegant. Commented Dec 15, 2015 at 15:00
• I was looking for telescoping but I could not find your blessed factor $\frac 12$ Commented Dec 15, 2015 at 15:02

Clearly $$U_{r+1}=\frac{r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$

$$2U_{r+1}=\frac{2r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$

$$2U_{r+1}=\frac{(2r+1)-1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$

$$2U_{r+1}=\frac{(2r+1)}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}$$

$$2U_{r+1}=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$

Now let $$V_r=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}$$

Then $$V_{r+1}=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$

Thus $$2U_{r+1}=V_r-V_{r+1}$$

$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=\sum_{r=1}^{n} \{V_r-V_{r+1}\}=V_1-V_{n+1}$$

$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=V_1-V_{n+1}$$

$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=\frac{1}{1}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}$$

• nice answer ..........+1 Commented Dec 15, 2015 at 15:12