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The sum of the infinite series $1+\frac{1}{2 !}+\frac{1.3}{4 !}+\frac{1.3 .5}{6 !}+\ldots \ldots$. is
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Verified Answer
The correct answer is:
$\sqrt{e}$
Hints : $\mathrm{T}_{\mathrm{n}}=\frac{1.3 .5 \ldots(2 n-1)}{\mid 2 n}$
$$
\begin{aligned}
& =\frac{\mid 2 n}{\lfloor 2 n(2.4 \ldots 2 n)} \\
& =\frac{\lfloor 2 n}{2^{\mathrm{n}}\lfloor n \mid 2 n} \\
& =\frac{x^{\mathrm{n}}}{\lfloor n} \\
& \therefore \frac{x}{\lfloor 1}+\frac{x^2}{\lfloor 2}+\ldots=e^x-1 \quad \frac{1}{2}=x \\
& \exp =1+e^x-1=e^x=e^{1 / 2}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{\mid 2 n}{\lfloor 2 n(2.4 \ldots 2 n)} \\
& =\frac{\lfloor 2 n}{2^{\mathrm{n}}\lfloor n \mid 2 n} \\
& =\frac{x^{\mathrm{n}}}{\lfloor n} \\
& \therefore \frac{x}{\lfloor 1}+\frac{x^2}{\lfloor 2}+\ldots=e^x-1 \quad \frac{1}{2}=x \\
& \exp =1+e^x-1=e^x=e^{1 / 2}
\end{aligned}
$$
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