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Okay so I'm asking this quesion knowing a thing or two about sequences and general terms

What is the sum of the series : $$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$

My Try: I tried calculating the general term $T_{n}$ for the sequence but I'm not able to understand how to include every single term of the sequence in a single general term. Writing the series again below the original series with the first term one ahead of the first term of the original series and subtraction didn't help too.How can I get this done? I get a feeling it could be a telescopic sum but unless I have a general term, I can't be sure of it.

Please suggest me any way of doing this. Thanks for giving this your time.

user362325
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Tanuj
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  • You have $1 + \frac{1.3}{6}+ \frac{1.3.5}{6.8} +...$. What does 1.3.5 mean? – Theo C. Aug 30 '17 at 03:13
  • @D.Beec the $.$ stands for multiplication, as in $1 \cdot 3 \cdot 5$ – Ben Grossmann Aug 30 '17 at 03:15
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    I don't what you're asking, they are in multiplication if that's what you're asking.And they do from an AP with first term 1 and common difference 2. – Tanuj Aug 30 '17 at 03:15
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    @Tanuj not every infinite sum has a nice answer (that is, there might not be a nice formula in terms of "$e,\pi$" or whatever you might expect). Is there a reason to expect a "nice" answer in this case? – Ben Grossmann Aug 30 '17 at 03:18
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    and the denominators? is it 6, 8, 10....? (the next multiplicand that is) – mdave16 Aug 30 '17 at 03:19
  • I have no idea what the next terms might be, and yes the answer does come out to be "nice". – Tanuj Aug 30 '17 at 03:20
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    @Tanuj To evaluate the sum, there has to be some sort of "rule" that defines what the terms are. How would one calculate the next term? – Carl Schildkraut Aug 30 '17 at 03:21
  • also, when you reveal new information in the comments, it can be useful to edit the question to reflect this piece of information. – mdave16 Aug 30 '17 at 03:22
  • @Carl Schildkraut That's what my question is! – Tanuj Aug 30 '17 at 03:22
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    @Tanuj Were you given this sequence, or did you come up with it? If it's the latter, how did you evaluate those first $3$ terms? – Carl Schildkraut Aug 30 '17 at 03:23
  • @Carl Schildkraut It's given in my book, just the same way I've typed it here. – Tanuj Aug 30 '17 at 03:25
  • perhaps it could help if you give us the sum value, then we could magic up a rule for what the terms should be? – mdave16 Aug 30 '17 at 03:28
  • @mdave if that's what you say, the sum is $4^\frac{2}{3}$. – Tanuj Aug 30 '17 at 03:29
  • Guys please give this your precious up votes so that it can attract the attention of others. It would really help. – Tanuj Aug 30 '17 at 03:31
  • easy, fourth term is $4^{2/3} - 1.8125$ and the fifth term is $0$, as is every term after that. – mdave16 Aug 30 '17 at 03:32
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    Yea, but thats not a really mathematical approach, plus you would've never gotten to the answer if you hadn't known it before. Also, how can every term after the fourth be zero? – Tanuj Aug 30 '17 at 03:34
  • how are you sure about the pattern of the numerators? I can tell you that if the numerator is the product of first n odds, the denominator is not 1, 6, 68, 68*10, .... – mdave16 Aug 30 '17 at 03:50
  • Maybe https://math.stackexchange.com/questions/32414/how-does-this-taylor-polynomial-work – Nosrati Aug 30 '17 at 03:52

2 Answers2

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HINT:

As the number of multiplicand in the numerator is greater than that of denominator and the common differences of numerator & denominator are both $2$

if $$S=1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$

$$\dfrac{-1\cdot S}{2\cdot4}=\dfrac{-1\cdot1}{2\cdot4}+\frac{-1\cdot1\cdot3}{2\cdot4\cdot6}+\frac{-1\cdot1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$$

Now utilize 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? $

lab bhattacharjee
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    Can you please expand your answer a bit more? It's a bit difficult to understand right now. – Tanuj Aug 30 '17 at 08:07
  • @Tanuj, See factorial starts with $1$ and we have $$6\cdot8=2^2(3\cdot4)$$ in the denominator. So we need to multiply with $$2\cdot4$$ to get $$2\cdot4\cdot6\cdot8=2^44!$$ The numerator should also have the same number of multiplicands. – lab bhattacharjee Aug 30 '17 at 09:17
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$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$$ Numerator is the product of odd positive integers which can be written as

$1 \cdot 3\cdot 5=\dfrac{1 \cdot 2\cdot 3\cdot 4 \cdot 5 }{2 \cdot 4}=\dfrac{5! }{2^2\left( 1\cdot 2\right)}=\dfrac{(2\cdot 2 +1)!}{2^2\cdot 2!}$

$n-$th numerator will be $\dfrac{(2n+1)!}{2^n \,n!}$

denominator is $6\cdot 8$ and can be written as $(2\cdot 3)(2\cdot 4)=2^2\cdot (3\cdot 4)=\dfrac{2^2\cdot 4!}{2}$

next denominator will be $6\cdot 8 \cdot 10=(2\cdot 3)(2\cdot 4)(2\cdot 5)=2^3(3\cdot 4\cdot 5)=\dfrac{2^3\cdot 5!}{2}$

$n-$th denominator will be $2^{n-1} (n+2)!$

Then the $n-$th term for $n\ge 2$ will be

$a_n=\dfrac{\dfrac{(2n+1)!}{2^n \,n!}}{2^{n-1} (n+2)!}=\dfrac{(2n+1)!}{2^n \,n!\,2^{n-1} (n+2)!}=\dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}$

And the series

$$1+\sum _{n=1}^{\infty } \dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}=4$$

Raffaele
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