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$\frac{1}{2 !}+\frac{1+2}{3 !}+\frac{1+2+3}{4 !}+\ldots$ is equal to :
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Verified Answer
The correct answer is:
$\frac{e}{2}$
We have $n$th term $T_n=\frac{1+2+3+\ldots+n}{(n+1) !}$
$=\frac{n(n+1)}{2 \cdot(n+1) !}=\frac{1}{2(n-1) !}$
Put $n=1,2,3, \ldots$
$T_1=\frac{1}{2}\left\lceil\frac{1}{0 !}\right\rceil, T_2=\frac{1}{2}\left\lfloor\frac{1}{1 !}\right\rfloor, T_3=\frac{1}{2}\left\lfloor\frac{1}{3 !}\right\rceil$
$\therefore T_1+T_2+T_3+\ldots=\frac{1}{2}\left[1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots\right]$
$=\frac{e}{2}$
$=\frac{n(n+1)}{2 \cdot(n+1) !}=\frac{1}{2(n-1) !}$
Put $n=1,2,3, \ldots$
$T_1=\frac{1}{2}\left\lceil\frac{1}{0 !}\right\rceil, T_2=\frac{1}{2}\left\lfloor\frac{1}{1 !}\right\rfloor, T_3=\frac{1}{2}\left\lfloor\frac{1}{3 !}\right\rceil$
$\therefore T_1+T_2+T_3+\ldots=\frac{1}{2}\left[1+1+\frac{1}{2 !}+\frac{1}{3 !}+\ldots\right]$
$=\frac{e}{2}$
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