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compute the summation

$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$

My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$

Now

\begin{align} & = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n} \\[10pt] & =\sum_ {n=1}^\infty \left(\frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}\right) \\[10pt] & =\sum_ {n=1}^\infty \left(\frac{1}{1\cdot2\cdot3\cdots(n-1)} -\frac{1}{2(1\cdot2\cdot3\cdots n}\right) \\[10pt] & =\sum_ {n=1}^\infty \left(\frac{1}{(n-1)!} - \frac{1}{2}\sum_ {n=2}^\infty \frac{1}{n!}\right) \\[10pt] & = e - \frac{1}{2} (e- 1)= \frac{1}{2}(e+1) \end{align}

Is it correct ???

if not correct then any hints/solution will be appreciated..

thanks in advance

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It is wrong since $$ 2\times 4\times 6\times\dotsb (2n)=2^n n!\quad (n\geq 1). $$ Write $$ \sum_{n=1}^\infty\frac{2n-1}{2^n n!}=\sum_{n=1}^\infty\frac{1}{2^{n-1}(n-1)!}-\sum_{n=1}^\infty\frac{(1/2)^n}{n!} =\sum_{n=0}^\infty\frac{(1/2)^n}{n!}-\sum_{n=1}^\infty \frac{(1/2)^n}{n!}=1. $$ Note that we don't need to know the value of the intermediate sums.

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    $\begingroup$ Why did you change $$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$ to $\sum_ {n=1}^{\infty} \frac{2n-1 }{2.4........(2n)}= ? \qquad$ $\endgroup$ – Michael Hardy Aug 4 '18 at 15:44
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For $n\ge1,$$$\dfrac{2n-1}{2^n n!}=\dfrac1{2^{n-1}(n-1)!}-\dfrac1{2^nn!}=f(n-1)-f(n)$$

where $f(m)=\dfrac1{2^mm!}$

Can you recognize the Telescoping nature and the surviving term(s) of the summation?

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Using telescopic approach with$$a_n=\dfrac{1}{2\cdot4\cdot6\cdot\cdots\cdot2n}$$we have $$S=\sum_{n=1}^{\infty}{a_n-a_{n-1}}=1$$

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You made a mistake after this line $$\sum_ {n=1}^{\infty}(\frac{2n}{2.4.6........2n} -\frac{1}{2.4.6........2n})$$ You simplified numerator and denominator by $2n$ but then $$\sum_ {n=1}^{\infty}(\frac{1}{2.4.6........2(n-1)} -\frac{1}{2.4.6........2n})$$

You have at the denominator $$2^nn!$$ So that $$\sum_ {n=1}^{\infty}\frac{2n}{2^nn!}=\sum_ {n=1}^{\infty}\frac{1}{2^{n-1}(n-1)!}$$

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