prove that 1/1^2 + 1/2^2 + 1/3^2 +............+1/n^2 < 2?

Question

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Answer 1

First observe that, for n>1,
n^2 > n(n-1).

Then 1/(n^2) < 1/(n(n-1).

Hence 1/2^2 < 1/(2*1), 1/3^2 < 1/(3*2), 1/4^2 < 1/(4*3), and so on.

Therefore 1/1^2 + 1/2^2 + 1/3^2 + ... + 1/n^2 <
1/1^2 + 1/(2*1) + 1/(3*2) + 1/(4*3) + ... + 1/(n*(n-1))

Noting that 1/(n*(n-1)) = 1/(n-1) - 1/n ,
we will have 1/(2*1) + 1/(3*2) + 1/(4*3) + ... + 1/(n*(n-1)) =
(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/(n-1) - 1/n) = 1 - 1/n

Therefore 1/1^2 + 1/2^2 + 1/3^2 + ... + 1/n^2 <
1/1^2 + 1/(2*1) + 1/(3*2) + 1/(4*3) + ... + 1/(n*(n-1)) =
1 + (1 - 1/n) = 2 - 1/n < 2.

Q.E.D

Answer 2

Most mathematics professors will not accept a "proof by picture"
A truly rigorous and formal proof follows:

1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...
is equal to:
1 + 1/(2*2) + 1/(3*3) + 1/(4*4) + ...
which is less than:
1 + 1/(1*2) + 1/(2*3) + 1/(3*4) + ...

Notice that,
1/2 = 1 - 1/2,
1/6 = 1/2 - 1/3,
1/12 = 1/3 - 1/4, ...

Thus 1 + 1/(1*2) + 1/(2*3) + 1/(3*4) + ...
is equal to:
1 + 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...
The halves, thirds, quarters and so on all balance out, so
1 + 1/(1*2) + 1/(2*3) + 1/(3*4) + ... = 2.

Hence 1 + 1/(2*2) + 1/(3*3) + 1/(4*4) < 2, qed.

You can check out:
http://en.wikibooks.org/wiki/Real_analysis/Series
for generalized details.

Feel free to email me for any further clarifications.

Answer 3

Draw the graph y=1/x².

Under the curve, you can draw the following histogram:
each bar is the rectangle bounded by y=0, y=1/n², x=n-1 and x=n.

You'll see that the histogram fits under the y=1/x²curve, therefore the area of the histogram is smaller than the area under the curve.

Let's leave aside the first bar (whose area is 1), and let's look at the area under the curve ffrom the second bar onwards (that's bounded by x=1 and x=infinity). This area is

integral(x=1 to infinity) 1/x²
= -1/x | (x=1, x=infinity)
= -1/inf + 1/1
= 1.

The area of th histogram = area of the first bar + area of the rest of the bars = 1 + (less that the area under the curve, to the right of x=1) = 1+1 = 2.

EDIT: Anyone who thumbs-down my so-called "proof by picture" doesn't know much about integrals, or just wants to be pedantic. The formula for my histogram is
f(x) = 1/n² where n=the smallest integer >= x.
It's easy to show that f(x)<=1/x², that there is strict inequality over some nonempty intervals and therefore that the sum of the series = the integral of f(x) < the integral of 1/x².

Answer 4

In case you're curious the harmonic series sums up to Pi^2/6