If Sn= 1.2/3!+2.22/4!+3.23/5!....upto nterms find lim n tends to infinity of Sn
Athinarayanan Ameraj , 11 Years ago
Grade 12
5 Answers
Arun Kumar
Last Activity: 10 Years ago
Hi Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
Arun Kumar
Last Activity: 10 Years ago
Hi little correction Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty
Koyena Basu
Last Activity: 8 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
Last Activity: 8 Years ago
jagdish singh singh
Last Activity: 8 Years ago
Sorry in last step , actually it is 1......................................................................
Provide a better Answer & Earn Cool Goodies
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free