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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?

mathreadler
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2 Answers2

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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 $1\cdot 3\cdot 5\cdot\ldots\cdot (2k+1)$.

Jack D'Aurizio
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  • 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. – Thomas Andrews Dec 15 '15 at 14:49
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    @Jack D'Aurizio strikes again :-)!Seeing you answer after quite some....and as usual you rocked it :-D!Thanks! –  Dec 15 '15 at 14:50
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    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. – Ron Gordon Dec 15 '15 at 15:00
  • I was looking for telescoping but I could not find your blessed factor $\frac 12$ – Piquito Dec 15 '15 at 15:02
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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 $$2U_{r+1}=V_r-V_{r+1}$$

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

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

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

Angelo Mark
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