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log(e)2=1-1/2+1/3-1/4+1/5-1/6+...oo पद त...

`log_(e)2=1-1/2+1/3-1/4+1/5-1/6+...oo` पद तक

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Observe the following lists {:("List-I","List-II"),((A) 1-(1)/(2)+(1)/(3)-(1)/(4)+(1)/(5)+...oo,(1)(2)/(3)+ (1)/(2)log2),((B)(1)/(5)+(1)/(2.5^(2))+(1)/(3.5^(3))+...oo,(2)log_(e )2),((C )(1)/(n+1)+(1)/(2(n+1)^(2))+(1)/(3(n+1)^(3))...oo,(3)-log_(e )((4)/(5))),((D)1+(1)/(3.3^(3))+(1)/(5.3^(5))+...oo,(4)log_(e )((5)/(4))),(,(5)-log_(e )(1-(1)/(n+1))):} The correct match for List - I from List -II is

The value of x satisfying the equation [3(1-(1)/(2)+(1)/(4).........oo)]^(log_(10)x)=[20(1-(1)/(4)+(1)/(16).........oo)]^(log_(x)1)

Show that, log_e2=1/(1.2)+1/(3.4)+1/(5.6)+.....infty

(1/(1!) +1/(2!) +1/(3!) + .....oo) (1/(2!) -1/(3!) +1/(4!)-1/(5!) .....oo)

If x = (1)/(1.2) + (1)/(3.4) + (1)/(5.6) …….oo, and y = 1 - (1)/(2.3) - (1)/(4.5) - (1)/(6.7)…….oo , then

If x = (1)/(1.2) + (1)/(3.4) + (1)/(5.6) …….oo, and y = 1 - (1)/(2.3) - (1)/(4.5) - (1)/(6.7)…….oo , then

(1+2/3+6/3^2+10/3^3+. . . + oo)^(log_(0.25)(1/3+1/3^2+1/3^3+. . . +oo)) is equal L and then find L^2

(1)/(2.3)+(1)/(4.5)+(1)/(6.7)+……oo=