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{:(" "Lt),(n rarr oo):}{ (1)/(2n+1)+(1)/...

`{:(" "Lt),(n rarr oo):}{ (1)/(2n+1)+(1)/(2n+2)+......+(1)/(3n)}=`

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The correct Answer is:
`log((3)/(2))`
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AAKASH SERIES-DEFINITE INTEGRALS-Exercise - 2.7 (Level-2)
  1. Lt(n rarr oo)[(1)/(n+1) + (1)/(n+2)+..(1)/(2n)]

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  2. Evaluate the limit . underset(n to 00)("lim") [(1)/(n+1)+(1)/(n+2)+…...

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  3. {:(" "Lt),(n rarr oo):}{ (1)/(2n+1)+(1)/(2n+2)+......+(1)/(3n)}=

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  4. Lt(n rarr oo)[(1)/(3n+1) + (1)/(3n+2)+…+(1)/(4n)]

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  5. Lt(n rarr oo)sum(r=0)^(n-1)((r)/(n^(2) + r^(2)))

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  6. Lt(ntooo)sum(r=1)^(n)(1)/(n)[sqrt ((n+r)/(n-r))]

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  7. Lt(n rarr oo) sum(r=1)^(n)[(1)/(sqrt(4n^(2) - r^(2)))]

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  8. Evaluate the limit. underset(n to 00)("lim") (sqrt(n+1)+sqrt(n+2)+…...

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  9. Lt(ntooo){(1)/(sqrt(n^(2)+1))+(1)/(sqrt(n^(2)+2^(2)))+.......+(1)/(sqr...

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  10. Lt(n rarr oo)[(1)/(n)+(1)/(sqrt(n^(2) -1^(2)))+(1)/(sqrt(n^(2)-2^(2)))...

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  11. Lt(n rarr oo)[(1n^(2))/((n+1)^(3))+(n^(2))/((n+2)^(3))+(n^(2))/((n+3))...

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  12. {:(" "Lt),(n rarr oo):} [(sqrt(n^(2)-1^(2)))/(n^(2))+(sqrt(n^(2)-2^(2)...

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  13. Lt(ntooo)(1)/(n)[sec^(2)""(pi)/(4n)+sec^(2)""(2pi)/(4n)+......+sec^(2)...

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  14. Lt(n rarr oo)[(n^(2))/((n^(2) +1)^(3//2))+(n^(2))/((n^(2) + 2)^(3//2))...

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  15. Lt(n rarr oo)[(sqrt(n))/(sqrt(n^(3)))+(sqrt(n))/(sqrt((n+4)^(3)))+.......

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  16. {:(" "Lt),(n rarr oo):} [ (1)/(n^(2)) sec^(2). (1)/(n^(2)) + (2)/(n^(2...

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  17. Lt(n rarr oo)[(1^(3))/(n^(4)+1^(4))+(2^(3))/(n^(4) + 2^(4))+....+(n^(3...

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  18. Lt(n rarr oo)(1)/(n)[e^((1)/(n))+e^((2)/(n))+....+e^((n)/(2))]

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  19. Lt(n rarr oo)[(1^(99)+2^(99)+3^(99)+...+n^(99))/(n^(100))]

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  20. Evaluate underset(n to oo)("lim")[(1+(1)/(n))(1+2/n)* * * (1+(n)/(n))...

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