Does telescoping works here?
n = 1 ∑ 1 0 ( n + 1 ) ( n + 2 ) n
Which of the following answer choices is the best estimate for the value of the sum above?
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1 solution
@Ankit Kumar Jain
We can write it as
S = ∑ n = 1 1 0 ( n + 1 ) ( n + 2 ) n
S = ∑ n = 1 1 0 ( n + 1 ) ( n + 2 ) 1 × ( n )
S = ∑ n = 1 1 0 ( n + 1 ) ( n + 2 ) ( n + 2 ) − ( n + 1 ) × ( n )
S = ∑ n = 1 1 0 n + 1 n − n + 2 n
S = 2 1 + 3 1 + 4 1 ⋯ + 1 1 1 − 1 2 1 0
@Tapas Mazumdar @abhishek alva @Chew-Seong Cheong Sir Can you pls tell a better approach?
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How did you evaluate the sum in that last step?
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It is just the sum of harmonic numbers.
H 1 0 − 1 2 1 0 − 1 which is equal to the required answer.
Note:- H n ≈ ln ( n ) + γ + 2 n 1 − 1 2 n 2 1
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@Rahil Sehgal – Oh...I didn't know that ..But what is that γ ?
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@Ankit Kumar Jain – Euler marshoni constant
ALthough the same ...you can look at it like this ( n + 1 ) ( n + 2 ) 2 ( n + 1 ) − ( n + 2 ) ...and then it easily breaks into partial fractions ..and then simplifying gives − 2 1 + 3 1 + 4 1 + 5 1 + 6 1 + 7 1 + 8 1 + 9 1 + 1 0 1 + 1 1 1 + 6 1 ...But how do we evaluate this?
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This expressions equals 1.187 as according to wolfram alpha but using the formula which you just told me gives value like 1.134...??How is that happening?
@Pi Han Goh @Jon Haussmann Sir , how is the problem meant to be done after simplification? I used Wolfram Alpha..Please explain me a proper solution..Thanks!