Question -4

Calculus Level 3

If f ( x ) = ∑ a = 1 n ( x − 1 a ) ⋅ ( x − 1 a + 1 ) f(x)=\displaystyle \sum_{a=1}^n \left( x-\frac{1}{a} \right) \cdot \left( x-\frac{1}{a+1} \right) , then find the value of lim ⁡ n → ∞ f ( 0 ) \displaystyle \lim_{n \to \infty} f(0)

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Vighnesh Raut
Feb 28, 2015

lim ⁡ n → ∞ f ( 0 ) = lim ⁡ n → ∞ ∑ a = 1 n ( − 1 a ) ( − 1 a + 1 ) = ∑ a = 1 ∞ 1 a ( a + 1 ) = ∑ a = 1 ∞ 1 a − 1 a + 1 = 1 − 1 2 + 1 2 − 1 3 + 1 3 − . . . . . . . . . . . . . ∞ = 1 \displaystyle{\lim _{ n\rightarrow \infty }{ f\left( 0 \right) } =\lim _{ n\rightarrow \infty }{ \sum _{ a=1 }^{ n }{ \left( -\frac { 1 }{ a } \right) } } \left( -\frac { 1 }{ a+1 } \right) =\sum _{ a=1 }^{ \infty }{ \frac { 1 }{ a(a+1) } } \\ =\quad \sum _{ a=1 }^{ \infty }{ \frac { 1 }{ a } -\frac { 1 }{ a+1 } } \\ =\quad 1-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } -\quad .............\infty =1}

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Use \ [ \ displaystyle{Enter Your latex code here} \ ] Don't Leave Space's (You know where) It Look's more good !

Deepanshu Gupta - 6 years, 3 months ago

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Ok....Thnx ...I will see to it from next time

Vighnesh Raut - 6 years, 3 months ago

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