Note that numerator is 1 less than 1 the corresponding denomerator for each term. Hence each term is 1 minus something which is positibe. Therefore the sum is less than 5.
Figure it out in general for x & n S(x,n) = x/(x+1) + (x+1)/(x+2) + .... + (x+k)/(x+k+1) +.... + (x+n-1)/(x+n) + (x+n)/x Expand each x/(x+i) S(x,n) = (1 - 1/(x+1) + 1 - 1/(x+2) + .... + 1 - 1/(x+k+1) + .... + 1 - 1/(x+n) + 1 + n/x Rearrange terms S(x,n) = (n+1) + (n/x - 1/(x+1) - 1(x+2) - ... - 1/(x+k+1) - .... - 1/(x+n) Break down n/x and rearrange terms S(x,n) = (n+1) + (1/x - 1/(x+1)) + (1/x - 1/(x+2)) + .... + (1/x - 1/(x+k+1)) + .... + (1/x - 1/(x+n)) Note 1/x > 1/(x+i) for i = 1 to n ∴ S(x,n) > (n+1) For this problem use x = 2021 and n = 4 S(x,n)/(n+1) converges downward to 1 as x → ∞ I am not sure yet if S(x,n)/(n+1) converges as n → ∞
It's easy to see that the sum is greater than 5. The first term in the sum is less than 1 by 1/2022, the second by 1/2023, the third by 1/2024, and the fourth by 1/2025. But the last term is greater than 1 by 4/2021. We can give one of those 4 to each of the first four terms, and 1/2021 is greater than all of the other fractions, so we gain back more than we lost. You only need seven or eight seconds to see this, without writing a single thing down.
Please see the video carefully then you have four fractions added with constant 5( it's only left hand side) afterthen you compare with rightside given constant 5
Rewrite as ... 1 - 1/2022 + 1 - 1/2023 + 1 - 1/2024 + 1 - 1/2025 + 1 + 4/2021 Now, since 1/2021 > 1/2022, 1/2021 > 1/2023, 1/2021 > 1/2024, 1/2021 > 1/2025, Then it follows that the quantity on the left is greater than 5.
SENSATIONAL!
Note that numerator is 1 less than 1 the corresponding denomerator for each term. Hence each term is 1 minus something which is positibe. Therefore the sum is less than 5.
Applying AM >= GM inequality for each of the fractions in the sum will get the result directly. (Note that the GM is 1)
💯 It's the quickest method.
5 - (1/2022 + 1/2023 .. + 1/2025) + 4/2021
Let x = (1/2022 + ... + 1/2025).
Since 1/2021 > 1/2022 > 1/2023 > .. > 1/2025, therefore 4/2021 > x.
Hence 5 - x + 4/2021 > 5.
Figure it out in general for x & n
S(x,n) = x/(x+1) + (x+1)/(x+2) + .... + (x+k)/(x+k+1) +.... + (x+n-1)/(x+n) + (x+n)/x
Expand each x/(x+i)
S(x,n) = (1 - 1/(x+1) + 1 - 1/(x+2) + .... + 1 - 1/(x+k+1) + .... + 1 - 1/(x+n) + 1 + n/x
Rearrange terms
S(x,n) = (n+1) + (n/x - 1/(x+1) - 1(x+2) - ... - 1/(x+k+1) - .... - 1/(x+n)
Break down n/x and rearrange terms
S(x,n) = (n+1) + (1/x - 1/(x+1)) + (1/x - 1/(x+2)) + .... + (1/x - 1/(x+k+1)) + .... + (1/x - 1/(x+n))
Note
1/x > 1/(x+i) for i = 1 to n
∴ S(x,n) > (n+1)
For this problem use x = 2021 and n = 4
S(x,n)/(n+1) converges downward to 1 as x → ∞
I am not sure yet if S(x,n)/(n+1) converges as n → ∞
5
It’s in my head.
Thanks for your beautiful feedback
It's easy to see that the sum is greater than 5. The first term in the sum is less than 1 by 1/2022, the second by 1/2023, the third by 1/2024, and the fourth by 1/2025. But the last term is greater than 1 by 4/2021. We can give one of those 4 to each of the first four terms, and 1/2021 is greater than all of the other fractions, so we gain back more than we lost. You only need seven or eight seconds to see this, without writing a single thing down.
Brilliant 👏
Yes .I noticed the same thing. It would be more of a challenge change the last term to 2230/2226
5 is more as the sum of the other is 4.448. only.
You only prove 4 fractions greater than 1, but not that their sum is greater than 5.
Please see the video carefully then you have four fractions added with constant 5( it's only left hand side) afterthen you compare with rightside given constant 5
😂
Rewrite as ...
1 - 1/2022 + 1 - 1/2023 + 1 - 1/2024 + 1 - 1/2025 + 1 + 4/2021
Now, since 1/2021 > 1/2022, 1/2021 > 1/2023, 1/2021 > 1/2024, 1/2021 > 1/2025,
Then it follows that the quantity on the left is greater than 5.
Overly complicated