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

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

Athinarayanan Ameraj , 13 Years ago

Grade 12

5 Answers

Arun Kumar

Hi
$\\S_n=\sum \frac{r*2^r}{(r+2)!} \\e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots \\\frac{e^x}{x}={1 \over x} + 1 + {x^2 \over x*2!} + {x^3 \over x*3!} + {x^4 \over x*4!} + \cdots \\ {d{e^x \over x} \over dx}={-1 \over x^2 }+0+S_n \\=>S_n={d{e^x \over x} \over dx}+{1 \over x^2 } \\=>S_n={e^x({{1 \over x}- {1 \over x^2}}})+{1 \over x^2 } \\=>S_n={e^2({{1 \over 2}- {1 \over 2^2}}})+{1 \over 2^2 }$
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Last Activity: 12 Years ago

Arun Kumar

Hi
little correction
$\\S_n=\sum \frac{r*2^r}{(r+2)!} \\e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots \\\frac{e^x}{x}={1 \over x} + 1 + {x^2 \over x*2!} + {x^3 \over x*3!} + {x^4 \over x*4!} + \cdots \\ {d{e^x \over x} \over dx}={-1 \over x^2 }+0+{1 \over 2!}+S_n \\=>S_n={d{e^x \over x} \over dx}+{1 \over x^2 }-{1 \over 2!} \\=>S_n={e^x({{1 \over x}- {1 \over x^2}}})+{1 \over x^2 }-{1 \over 2!} \\=>S_n={e^2({{1 \over 2}- {1 \over 2^2}}})+{1 \over 2^2 }-{1 \over 2!}$
Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Last Activity: 12 Years ago

Koyena Basu

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

Last Activity: 9 Years ago

jagdish singh singh

$\hspace{-0.70 cm}$ 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}.$$