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Sum to Infinity of r·2^r/(r+2)! Series | JEE

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Question
If Sn=1⋅23!+2⋅224!+3⋅235!+⋯S_n=\frac{1\cdot2}{3!}+\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\cdots up to nn terms, then the sum to infinite terms is
A4π\frac{4}{\pi}
B3e\frac{3}{e}
Cπ3\frac{\pi}{3}
D11correct
Solution
Step 1: General term tr=r⋅2r(r+2)!t_r=\dfrac{r\cdot 2^r}{(r+2)!}. Write r=(r+2)−2r=(r+2)-2:
tr=(r+2)2r−2⋅2r(r+2)!=2r(r+1)!−2r+1(r+2)!.t_r=\frac{(r+2)2^r-2\cdot 2^r}{(r+2)!}=\frac{2^r}{(r+1)!}-\frac{2^{r+1}}{(r+2)!}.
 Step 2: Sum from r=1r=1 to nn; it telescopes:
Sn=212!−2n+1(n+2)!=1−2n+1(n+2)!.S_n=\frac{2^1}{2!}-\frac{2^{n+1}}{(n+2)!}=1-\frac{2^{n+1}}{(n+2)!}.
 Step 3: As n→∞n\to\infty the second term →0\to 0, so S∞=1.S_\infty=1. Correct answer: (4)
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