Proving that $sumlimits_{n=0}^{infty }frac{1}{(2n)!!}=sqrt{e}$
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
246k23221455
add a comment |
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
246k23221455
1
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
– Lucian
Jan 16 '15 at 19:18
add a comment |
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
246k23221455
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
246k23221455
edited Jan 3 at 20:46
Did
246k23221455
edited Jan 3 at 20:46
Did
246k23221455
edited Jan 3 at 20:46
Did
246k23221455
edited Jan 3 at 20:46
Did
246k23221455
246k23221455
asked Jan 16 '15 at 18:35
user187581
1
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
– Lucian
Jan 16 '15 at 19:18
add a comment |
1
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
– Lucian
Jan 16 '15 at 19:18
1
1
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
– Lucian
Jan 16 '15 at 19:18
Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
– Lucian
Jan 16 '15 at 19:18
add a comment |
2 Answers
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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
Crostul
27.6k22352
add a comment |
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
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 Etincelle
12.3k12147
add a comment |
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2 Answers
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2 Answers
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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
Crostul
27.6k22352
add a comment |
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
Crostul
27.6k22352
add a comment |
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
Crostul
27.6k22352
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
Crostul
27.6k22352
answered Jan 16 '15 at 18:38
Crostul
27.6k22352
answered Jan 16 '15 at 18:38
Crostul
27.6k22352
answered Jan 16 '15 at 18:38
Crostul
27.6k22352
27.6k22352
add a comment |
add a comment |
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
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 Etincelle
12.3k12147
add a comment |
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
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 Etincelle
12.3k12147
add a comment |
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
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 Etincelle
12.3k12147
$$e^x = sum_{n=0}^inftydfrac{x^n}{n!}$$
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 Etincelle
12.3k12147
answered Jan 16 '15 at 18:38
Petite Etincelle
12.3k12147
answered Jan 16 '15 at 18:38
Petite Etincelle
12.3k12147
answered Jan 16 '15 at 18:38
Petite Etincelle
12.3k12147
12.3k12147
add a comment |
add a comment |
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Could you also prove that $quaddisplaystylesum_{n=0}^inftyfrac1{(2n+1)!!}=sqrt e~int_0^1e^{-x^2/2}~dxquad?~$ :-)
– Lucian
Jan 16 '15 at 19:18