(1/(1!) +1/(2!) +1(3!) + .....oo) (1/(2!...

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

`1/e`

`1-1/e`

`e-1`

`1/e -1`

Verified by Experts

The correct Answer is:

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1+(log(e)5)/(1!) +(loge5)^2/(2!)+(loge5)^3/(3!)+ .......oo =
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1-log2+((log2)^2)/(2!) -(log2)^3/(3!)+....oo
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1+ 2 (logeq)+(2^2)/(2!) (logea)^2 +(2^3)/(3!) (loge a)^3+...=
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(1/(1!) +1/(2!) +1(3!) + .....oo) (1/(2!) +1/(3!) +1(4!)-1/(5!) .....o...
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(1-1/(1!) +9/(2!) +(27)/(3!) + .....oo) (1+9/(1!) +(81)/(2!)+(729)/(3!...
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sum(n=1) ^(oo) (1)/((2n-1)!)=
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e^(7x)-e^(-7x)=
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The 12th term in the expansion of e^(-3) is
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Coffiecient of x^3 in the expansion of e^(-bx)
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A : The value of e lies between 2 and 3. R : e=1+(1)/(1!)+(1)/(2!)+(...
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