Prove that
one over one squared plus one over two squared plus one over three squared....... plus one over 1000 squared is smaller than two. Details please.
1 * 1 = 1
1/3 * 1/3 = 1/9
1/1000 * 1/1000 = 1/1,000,000
Here are three proofs, take your pick.
http://en.wikipedia.org/wiki/Basel_problem
1/2^2<1/(1*2)=1-1/2
1/3^2<1/(2*3)=1/2-1/3
1/4^2<1/(3*4)=1/3-1/4
......................
1/999^2<1/(998*999)=1/998-1/999
1/1000^2<1/(999*1000)=1/999-1/1000
Now add all inequalities
http://www.southernct.edu/~sandifer/Ed/History/Preprints/Talks/NYU%20Basel%20Problem%20Paper.PDF
Could you expand more on the answer anoymomous?
To prove that the expression "1/1^2 + 1/2^2 + 1/3^2 + ... + 1/1000^2" is less than 2, we can use an approach called the comparison test.
Let's consider the harmonic series, which is the sum of the reciprocal of the positive integers: 1 + 1/2 + 1/3 + 1/4 + ...
The comparison test states that if a series is term-wise less than another series which is known to diverge, then the original series also diverges. In this case, since the harmonic series diverges, we can prove that the given series also diverges.
First, let's rewrite the given series as "1/1^2 + 1/2^2 + 1/3^2 + ... + 1/n^2 + ...". Note that "n" can be any positive integer.
Now, let's consider the sum of the reciprocal of squares starting from 1/4 instead of 1/1: 1/4 + 1/5 + 1/6 + ... + 1/n^2 + ...
We can see that this series is term-wise less than the harmonic series because each term is a reciprocal of a larger number. In other words, each term in the new series is less than or equal to the corresponding term in the harmonic series.
Since the harmonic series diverges, this new series also diverges. Therefore, the original series "1/1^2 + 1/2^2 + 1/3^2 + ... + 1/1000^2" must also diverge.
Now, we need to determine the upper limit of this diverging series. We can use the integral test to estimate the sum.
The integral test states that if a function f(x) is positive, continuous, and decreasing for x ≥ 1, and if the series "f(1) + f(2) + f(3) + ... + f(n) + ..." and the integral "∫f(x) dx" both exist, then the series and the integral either both converge or both diverge.
In this case, let f(x) = 1/x^2. We can see that f(x) is positive, continuous, and decreasing for x ≥ 1.
Now, let's integrate f(x) from 1 to infinity:
∫(1/x^2) dx = [-1/x] evaluated from 1 to infinity = (-1/infinity) - (-1/1) = 0 + 1 = 1
Since the integral of f(x) is equal to 1, which is a finite value, this implies that the original series converges.
Therefore, the sum of "1/1^2 + 1/2^2 + 1/3^2 + ... + 1/1000^2" is less than the upper limit of a converging series, which is 1.
Since 1 is less than 2, we can conclude that "1/1^2 + 1/2^2 + 1/3^2 + ... + 1/1000^2" is smaller than 2.