Series question,related to telescopic series, 1/24+ 13/246+ 135/246*8 ...infinity

Question

The series is $$\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....$$

It continues to infinity.I tried multiplying with $2$ and dividing each term by$(3-1)$,$(5-3)$ etc,starting from the second term which gives me $$\frac{1}{8} -\frac{1}{4*6} -\frac{1}{4*6*8}-\frac{1}{4*6*8*10}$$.

Also if anybody is wondering the answer 0.5

Are you meaning to say the terms tothe left of each / sign is in the numerator and the terms to the right are in the denominator? Are you trying to prove convergence or divergence? — Alan, Oct 02 '14 at 04:11
as a likely hint, try putting the series into general form, then using the ratio test — Alan, Oct 02 '14 at 04:13
@Shreyas, Where have you found "Related to telescopic Series"? — lab bhattacharjee, Oct 02 '14 at 04:33
This question comes under the topic "Telescopic series" so using that method is probably the fastest method. — Shreyas, Oct 02 '14 at 04:42
This seems to be the same question as: http://math.stackexchange.com/questions/936236/sum-of-the-series-frac12-cdot-4-frac1-cdot32-cdot4-cdot6-dots — Martin Sleziak, Oct 02 '14 at 19:03

Olivier Oloa · Answer 1 · 2014-10-02T18:25:00.973

2

We have $$ \begin{align} \frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}&=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\\\ &=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\\\ & =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2n+2)}\\\\ &=\frac{1}{(2n+2)\,4^n}\binom{2n}{n}\\\\&=\frac{1}{4^n}\binom{2n}{n}-\frac{1}{4^{n+1}}\binom{2n+2}{n+1}. \end{align} $$ The series is then telescopic: $$\sum_{n=1}^{N}\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}=\frac{1}{2}-\frac{1}{4^{N+1}}\binom{2N+2}{N+1}. $$As $N$ is great, one may prove that $$\frac{1}{4^{N+1}}\binom{2N+2}{N+1}=O\left(\frac{1}{\sqrt{N}}\right). $$ The series is thus equal to $\dfrac{1}{2}$:

$$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}=\frac{1}{2}. $$

edited Oct 02 '14 at 18:25

answered Oct 02 '14 at 04:43

Olivier Oloa

120,989

I'm really sorry but can u tell me which notation you are using in the bracket so i may look it up. – Shreyas Oct 02 '14 at 04:47
@Shreyas You mean notation for $\binom{2n}{n}$? It is: \binom{2n}{n}. – Olivier Oloa Oct 02 '14 at 04:49
yes thank you . – Shreyas Oct 02 '14 at 04:50
@OlivierOloa, So, the sum should be $\dfrac12+\dfrac12?$ – lab bhattacharjee Oct 02 '14 at 05:34
@labbhattacharjee I've checked my result: the sum is equal to $\dfrac{1}{2}$. Please, check it with a CAS. Thank you. – Olivier Oloa Oct 02 '14 at 18:32
@OlivierOloa, Can you please pinpoint the mistake in my answer? – lab bhattacharjee Oct 02 '14 at 18:34
@labbhattacharjee Dear lab bhattacharjee, please I don't really understand the reasoning in your answer. – Olivier Oloa Oct 02 '14 at 18:42
@OlivierOloa, Please have a look into the linked answer,too if you patience & time – lab bhattacharjee Oct 02 '14 at 18:44
@labbhattacharjee Please, have a look into the hyperlink in the OP question, I think your mistake comes from the fact you sum the series from $0$ index instead of starting it from $1$ index. – Olivier Oloa Oct 04 '14 at 06:47
@OlivierOloa, Thanks for your feedback – lab bhattacharjee Oct 04 '14 at 06:54

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

The $r$th term $\displaystyle\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$ where integer $r\ge1$

As the number of terms in the numerator $=$ the number of terms in the denominator $-1,$

If $\displaystyle S=\sum_{n=1}^\infty\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)},$

Multiplying the numerator by the previous term of $1,3,5,\cdots$ (Arithmetic Series)

$\displaystyle -S=\sum_{n=1}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$

$\displaystyle=-\frac{-1}2+\sum_{n=0}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}$

Now, $\displaystyle\sum_{n=0}^\infty\frac{(-1)1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n+2)}=(1-1)^{\frac12}-1$

Now follow Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $

Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity

2 Answers2

Series question,related to telescopic series, 1/24+ 13/246+ 135/246*8 ...infinity