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If S n = 1.2/3!+2.2 2 /4!+3.2 3 /5!....upto nterms find lim n tends to infinity of S n

If Sn= 1.2/3!+2.22/4!+3.23/5!....upto nterms find lim n tends to infinity of Sn 

Grade:12

5 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
7 years ago
Hi

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
Arun Kumar IIT Delhi
askIITians Faculty 256 Points
7 years ago
Hi
little correction

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
Koyena Basu
12 Points
4 years ago
The correct answer is 1e^x = 1 + x + x^2/2! + x^3/3! + ...Dividing by x^2:e^x/x^2=1/x^2 + 1/x + 1/2! + x/3! +...Differentiating above expression and then multiplying the same by x, we get :-2x/x^3 - x/x^2 + 0 + Sn Thus, Sn = x* d(e^x/x^2)/dx + 2/x^2 + 1/xSn= 1/2+1/2+ 2e^x(x-2)/x^3Thus Sn = 1+0=1Hope it helps
jagdish singh singh
173 Points
4 years ago
\hspace{-0.70 cm}$ Let $S_{n} = \lim_{n\rightarrow \infty}\sum^{n}_{r=0}\frac{r\cdot 2^r}{(r+2)!} = \lim_{n\rightarrow \infty}\frac{1}{2}\sum^{n}_{r=0}\bigg[\frac{((r+2)-2)\cdot 2^{r+1}}{(r+2)!}\bigg]$\\\\\\ So $\displaystyle S_{n}=\lim_{n\rightarrow \infty}\frac{1}{2}\sum^{n}_{r=1}\left[\frac{2^{r+1}}{(r+1)!}-\frac{2^{r+2}}{(r+2)!}\right]$\\\\ So Using Telescoping Sum, We get $\displaystyle S_{n} = \lim_{n\rightarrow \infty}\frac{1}{2}\bigg[1-\frac{2^{n+2}}{(n+2)!}\bigg] = \frac{1}{2}.$
jagdish singh singh
173 Points
4 years ago
Sorry in last step , actually it is 1......................................................................
 

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