n → ∞ lim n 1 + n 2 + n 2 2 + n 2 3 + ⋯ + n 2 n
Find the limit to 2 decimal places.
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Can you clarify what you mean by "AP formula"? I believe that you're using something differnt from arithmetic progression sum . I think what you want it:
( 1 − x ) ( 1 + x + x 2 + … + x n ) = ( 1 − x n + 1 ) .
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I meant GP sum formula, my bad
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Ah ic. I've edited that into the solution.
In future, you can edit your solution directly by hitting the "edit" button at the bottom of the solution.
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@Calvin Lin – I was trying to, and I did it before with other solutions of other problems, but my browser did not let me save the editions
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@Guilherme Niedu – Hm, can you elaborate on "browser did not let me save the editions"? What happened then?
@Calvin Lin – Thanks anyway!
Relevant wiki: Riemann Sums
L = n → ∞ lim n 1 + n 2 + n 2 2 + n 2 3 + . . . + n 2 n = n → ∞ lim n 1 k = 0 ∑ n 2 n k = ∫ a b 2 x d x = ∫ 0 1 2 x d x = ln 2 2 x ∣ ∣ ∣ ∣ 0 1 = ln 2 2 − 1 = ln 2 1 ≈ 1 . 4 4 By Riemann sums a = n → ∞ lim n 0 = 0 , b = n → ∞ lim n n = 1
The limit of summation should be from k=1 to k=n or k=0 to k-n-1.
I think first of all 1/n should be separated which will become zero.then rest of the things will be same as above
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L = n → ∞ lim n 1 + n 2 + n 2 2 + n 2 3 + . . . + n 2 n = n → ∞ lim n 2 n 0 + 2 n 1 + 2 n 2 + 2 n 3 + . . . + 2 n n = n → ∞ lim n 1 k = 0 ∑ n 2 n k
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But the limit of summation should be k=1 to k=n
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@Kushal Bose – What do you mean the limit of summation? I am quite sure my solution is correct. Refer to Riemann sums.
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@Chew-Seong Cheong – Your solution is correct but I am saying about lower limit and upper limit of summation that is it should be k=1 to k=n
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@Kushal Bose – No, I have shown that 1 = 2 n 0 , therefore the limit s are k = 0 to k = n .
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@Chew-Seong Cheong – But riemann sum is valid for k=1 to k=n or k=0 to k=n-1
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@Kushal Bose – No, I have shown that ∫ a b 2 x d x . a = lim n → ∞ n 0 and b = lim n → ∞ n n . The limits of integration follow those of summation. Refer to the wiki I have attached. I learned from it.
Even if k = 1 to k = n − 1 , still a = 0 and b = 1 because n → ∞ .
You are everywhere 😂
You come up with different but very good solutions , so I guess it's good .👍
And one more thing your handwriting is really beautiful!
Did the same! . the infamous Riemann's sum technique
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L = n → ∞ lim n 1 + n 2 + n 2 2 + n 2 3 + ⋯ + n 2 n
By the geometric progression sum formula,
L = n → ∞ lim n ( 2 1 / n − 1 ) 2 1 + 1 / n − 1
Set m = 1 / n , we get
L = m → 0 lim 2 m − 1 m ( 2 1 + m − 1 ) = ( m → 0 lim 2 1 + m − 1 ) ⋅ ( m → 0 lim 2 m − 1 m ) = m → 0 lim 2 m − 1 m
Applying L'Hôpital, we get
L = m → 0 lim 2 m ⋅ l n ( 2 ) 1 = l n ( 2 ) 1 = 1 . 4 4 . . .