#Comparison of inequalities for series
naive hollowBOT
dusty summit
How so?
steady stream
How so?
So what my rational behind the first is that 1/(n^2 + sin(n)) =< 1/n^2. Then by p-series we can say it is convergent
steady stream
How so?
but i don’t understand because 1/n^2 - sin(n) also has that same logic where you can compare it with 1/n^2
dusty summit
The problem is sin(n) oscillates between -1 and 1
So the comparison fails
n² > n² + sin(n)
0 > sin(n) which is not for all n true
Even for very big n
You can take 0.5n²
0.5n² < n² + sin(n)
0.5n² > -sin(n) true
steady stream
0.5n² < n² + sin(n)
0.5n² > -sin(n) true
1/n^2 - sin(n) is comparable to 1/n^2 is valid? But doesn’t sin(n) also bop around -1 and 1?
steady stream
0.5n² < n² + sin(n)
0.5n² > -sin(n) true
Then by this, would n > n-1 also be true, making that comparison valid?
dusty summit
1/n^2 - sin(n) is comparable to 1/n^2 is valid? But doesn’t sin(n) also bop arou...
This was my exact point
Hence why I said 0.5n² should work
dusty summit
0.5n² < n² + sin(n)
0.5n² > -sin(n) true
.
dusty summit
Then by this, would n > n-1 also be true, making that comparison valid?
No, if you noticed the 3rd one are the exact same series, they didnt use comparison test but just shifted by one index