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Proving that $sumlimits_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$
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Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
sequences-and-series
edited Jan 3 at 20:46
Did
247k23222458
$endgroup$
add a comment |
$begingroup$
Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
sequences-and-series
edited Jan 3 at 20:46
Did
247k23222458
$endgroup$
1
$begingroup$
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
$endgroup$
– Lucian
Jan 16 '15 at 19:18
add a comment |
$begingroup$
Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
sequences-and-series
edited Jan 3 at 20:46
Did
247k23222458
$endgroup$
Proving that $$sum_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$$
Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
sequences-and-series
sequences-and-series
edited Jan 3 at 20:46
Did
247k23222458
edited Jan 3 at 20:46
Did
247k23222458
edited Jan 3 at 20:46
Did
247k23222458
edited Jan 3 at 20:46
Did
247k23222458
edited Jan 3 at 20:46
Did
247k23222458
247k23222458
asked Jan 16 '15 at 18:35
user187581
1
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Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
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– Lucian
Jan 16 '15 at 19:18
add a comment |
1
$begingroup$
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
$endgroup$
– Lucian
Jan 16 '15 at 19:18
1
1
$begingroup$
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
$endgroup$
– Lucian
Jan 16 '15 at 19:18
$begingroup$
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
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– Lucian
Jan 16 '15 at 19:18
add a comment |
2 Answers
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$begingroup$
Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
so your series is just
$$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
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$begingroup$
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
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2 Answers
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2 Answers
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$begingroup$
Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
so your series is just
$$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
$endgroup$
add a comment |
$begingroup$
Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
so your series is just
$$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
$endgroup$
add a comment |
$begingroup$
Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
so your series is just
$$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
$endgroup$
Note that $$(2n)!! = 2cdot4cdot 6 cdots 2n = 2^n n!$$
so your series is just
$$sum_n frac{(1/2)^n}{n!} = e^{frac{1}{2}}$$
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
answered Jan 16 '15 at 18:38
CrostulCrostul
27.8k22352
27.8k22352
add a comment |
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$begingroup$
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
add a comment |
$begingroup$
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
add a comment |
$begingroup$
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
Plug $x = dfrac{1}{2}$, then $dfrac{x^n}{n!} = dfrac{1}{2^n n!} = dfrac{1}{2 cdot 4 cdot 6 cdots 2n} = dfrac{1}{(2n)!!}$
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
answered Jan 16 '15 at 18:38
Petite EtincellePetite Etincelle
12.4k12148
12.4k12148
add a comment |
add a comment |
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Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
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– Lucian
Jan 16 '15 at 19:18