Just Sum Up!
n → ∞ lim ( n 1 + ( n + 1 ) 3 n 2 + ( n + 2 ) 3 n 2 + ⋯ + 8 n 1 ) = ?
The answer is 0.375.
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1 solution
The lower limit in the summation should be r = 1
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If r = 0 ⟹ n 1 . So its correct.
No, it's correct.
Dude Its correct.
Ha ha, it ain't gonna crush that easily. :D.
As a matter of interest, I let Wolfram-alpha do its thing. It didn't like the limit approach, so I let n = 10^20, and evaluated sum of n^2/(n+r)^3 from r=0 to r=10^20. Actually, I started with less zero,s and then increased, but the strangest thing is that it insisted on digits remaining after the .375. To be more specific, at n = 10^20 it gave an answer of ,375 followed by 17 zeroes, then 5625 followed by 17 zeroes, then 234375, Actaully 2343749999999999999999.......... Interesting. These numbers were present for other values of n
n → ∞ lim [ n 1 + ( n + 1 ) 3 n 2 + ( n + 2 ) 3 n 2 + . . . . . . + 8 n 1 ] = n → ∞ lim r = 0 ∑ n ( n + r ) 3 n 2 = n → ∞ lim r = 0 ∑ n n 1 × ( 1 + n r ) 3 1 = ∫ 0 1 ( 1 + x ) 3 d x = − [ 2 ( 1 + x ) 2 1 ] 0 1 = − 2 1 × ( 4 1 − 1 ) = 8 3 = 0 . 3 7 5 .