Home
Class 12
MATHS
Let Sn=1+2+3++n and Pn=(S2)/(S2-1)dot(S...

Let `S_n=1+2+3++n` and `P_n=(S_2)/(S_2-1)dot(S_3)/(S_3-1)dot(S_4)/(S_4-1)...(S_n)/(S_n-1)` Where `n in N ,(ngeq2)dot` Then `("lim")_(n→oo)P_n=______`

Text Solution

Verified by Experts

The correct Answer is:
`:.lim_(nto oo)P_(n)=lim_(ntooo)(3n)/(n+2)=3`
Promotional Banner

Topper's Solved these Questions

  • LIMITS

    ARIHANT MATHS ENGLISH|Exercise Exercise (Single Option Correct Type Questions)|39 Videos
  • LIMITS

    ARIHANT MATHS ENGLISH|Exercise Exercise (More Than One Correct Option Type Questions)|15 Videos
  • INVERSE TRIGONOMETRIC FUNCTIONS

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|8 Videos
  • LOGARITHM AND THEIR PROPERTIES

    ARIHANT MATHS ENGLISH|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|2 Videos

Similar Questions

Explore conceptually related problems

Let S_(n)=1+2+3+...+n " and " P_(n)=(S_(2))/(S_(2)-1).(S_(3))/(S_(3)-1).(S_(4))/(S_(4)-1)...(S_(n))/(S_(n)-1) , where n inN,(nge2) Then underset(ntooo)limP_(n) =__________.

If s_n=(1-4/1)(1-4/9)(1-4/25)......(1-4/((2n-1)^2)), where n in N, then

Let S_n=1/1^2 + 1/2^2 + 1/3^2 +….. + 1/n^2 and T_n=2 -1/n , then :

If S_(n)=(1^(2).(2))/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(3!)+…(n^(2).(n+1))/(n!) then lim_(n rarr infty) S_(n) is equal to

The value of lim_(n->oo)[(2n)/(2n^2-1)cos(n+1)/(2n-1)-n/(1-2n)dot(n(-1)^n)/(n^2+1)]i s 1 (b) -1 (c) 0 (d) none of these

If S_(n)=(1^(2)-1+1)(1!)+(2^(2)-2+1)(2!)+...+(n^(2)-n+1)(n!) , then S_(50)=

If S_(1), S_(2), S_(3),...,S_(n) are the sums of infinite geometric series, whose first terms are 1, 2, 3,.., n and whose common rations are (1)/(2), (1)/(3), (1)/(4),..., (1)/(n+1) respectively, then find the values of S_(1)^(2) + S_(2)^(2) + S_(3)^(2) + ...+ S_(2n-1)^(2) .

If S_(n)=1+1/2+1/3+…+1/n(ninN) , then prove that S_(1)+S_(2)+..+S_((n-1))=(nS((n))-n)or(nS((n-1))-n+1)

Let S_n=Sigma_(k=1)^(4n) (-1)^((k(k+1))/2)k^2 .Then S_n can take value (s)

If S_n denotes the sum of n terms of A.P., then S_(n+3)-3S_(n+2)+3S_(n+1)-S_n= (a) S_2-n b. S_(n+1) c. 3S_n d. 0