prove why sum(1 to inf) of 1/(n^2) is less than 2?

Question

Can someone please explain how prove this statement?

sum (1 to infinity) of 1/(n^2) is less than 2

Why is it less than 2? Please give a detailed explaination if you can.

Answer 1

First off,

1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ......

is less than

1 + 1/2 + 1/4 + 1/8 + 1/16 ...

after the first term. That is, although the first terms are the same, each successive term of the first series is less than the corresponding term of the second series. Therefore

1 <= 1
1/2^2 < 1/2
1/3^2 < 1/4
1/4^2 < 1/8
1/5^2 < 1/16

And the second series represents an infinite geometric series with r = 1/2 and a = 1. The sum of an infinite geometric series is equal to

S = a/(1 - r), so

S = 1/(1 - (1/2))
S = 1/[1/2]
S = 2

Which means

[1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... ] < [ 1 + 1/2 + 1/4 + 1/8 + 1/16 ... ]

But the sum on the right hand side is equal to 2, so

1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... < 2

Answer 2

This is called the Basel problem. The solution others are offering are hardly the most insightful, nor the most rigorous. For more on this problem, see:
http://en.wikipedia.org/wiki/Basel_problem

We make use of the fact that:
1/n² < 1 / (n × (n+1) )

to show:
Σ(1 to N) 1/n² = 1 + Σ(2 to N) 1/n² < 1 + Σ(2 to N) 1/(n (n+1) )

Let's look at the sum on the far right:
Σ(2 to N) 1/(n (n+1) ) =
Σ(2 to N) 1/(n-1) - 1/n

That is a telescoping series. See:
http://en.wikipedia.org/wiki/Telescoping_series

It sums to:
Σ(2 to N) 1/(n-1) - 1/n = 1 - 1/N

Thus:
Σ(1 to N) 1/n² ≤ 1 + Σ(2 to N) 1/(n (n+1) ) = 2 - 1/N

And finally:
Σ(1 to ∞) 1/n² < lim(N→∞) 2 - 1/N = 2

Answer 3

The reason why the sum of your equation when run one to infinity is less then 2 is because your devisor (the number under the division bar) becomes exponentially greater you have a smaller and smaller number to add, so your sum would look something like 1 + 1/4 + 1/16 etc so the final answer as you can see will never grow large enough to pass the number 2.

Answer 4

For an explanation of the p series:

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

As quoted from wikipedia:
BEGIN QUOTE
"
"p-series"
The p-series, is (any of) the series


for p a positive real number. The series is always convergent if p > 1 and divergent otherwise. When p = 1, the series is the harmonic series. If p > 1 then the sum of the series is ζ(p), i.e., the Riemann zeta function evaluated at p.

"
END QUOTE from wikipedia


For a proof of convergence of the p series:

http://64.233.167.104/search?q=cache:zNlrCgehDhoJ:www.math-cs.cmsu.edu/~mjms/2004.2/rkhan.ps+prove+p+series+converges&hl=en&ct=clnk&cd=7&gl=us&ie=UTF-8


What I would recommend is that you look up in a book if you have one, "p Series" or "The Ratio Test" .

-----------------------------------------------------------------------------
BELOW THIS LINE IS MY SPECULATION, NOT ACTUAL ADVICE....


I would want to use the infinite series sum from n=1 to infinity of 1/(2^(n-1)). This is known to have a limit of 2.
see this link(the proof is not there though)
http://www.math.utah.edu/~carlson/teaching/calculus/series.html



to continue,
I write this on a piece of paper:
*THEIRS*
sum from 1 to infinity of 1/(2^(n-1))
S1 = 1
S2 = 1 + 1/2 = 3/2 = 1.5
S3 = 1 + 1/2 + 1/4 = 7/4 = 1.75
S4 = 1 + 1/2 + 1/4 + 1/8 = 15/8 = 1.875

the nth partial sum is
Sn = 2-(1/2)^n

so as n approaches infinity, Sn approaches 2.
*OURS*
But for our series,
sum from n = 1 to infinity of 1/(n^2)
S1 = 1
S2 = 1 + 1/4 = 1.25
S3 = 1 + 1/4 + 1/9 = 1.36111
S4 = 1 + 1/4 + 1/9 + 1/16 = 1.42361

p is greater than 1 so we know it is convergent.
but to prove the infinite series of this converges to LESS THAN 2, we have to show something. compare it to the other series above?

Theirs vs Ours
Sn, n=1: 1 vs 1
Sn, n=2: 1.5 vs 1.25
Sn, n=3: 1.75 vs 1.36111111111111
Sn, n=4: 1.875 vs 1.42361111111111
Sn, n=5: 1.9375 vs 1.46361111111111
Sn, n=6: 1.96875 vs 1.49138888888889
Sn, n=7: 1.984375 vs 1.5117970521542
Sn, n=8: 1.9921875 vs 1.5274220521542
Sn, n=9: 1.99609375 vs 1.53976773116654
Sn, n=10: 1.998046875 vs 1.54976773116654
Sn, n=11: 1.9990234375 vs 1.55803219397646
Sn, n=12: 1.99951171875 vs 1.5649766384209
Sn, n=13: 1.999755859375 vs 1.57089379818422
Sn, n=14: 1.9998779296875 vs 1.57599583900054
Sn, n=15: 1.99993896484375 vs 1.58044028344499
Sn, n=16: 1.99996948242187 vs 1.58434653344499
Sn, n=17: 1.99998474121093 vs 1.58780674105744
Sn, n=18: 1.99999237060546 vs 1.59089316081053
Sn, n=19: 1.99999618530273 vs 1.59366324391302
Sn, n=20: 1.99999809265136 vs 1.59616324391302
Sn, n=21: 1.99999904632568 vs 1.59843081760917
Sn, n=22: 1.99999952316284 vs 1.60049693331165
Sn, n=23: 1.99999976158142 vs 1.60238729247989
Sn, n=24: 1.99999988079071 vs 1.604123403591
Sn, n=25: 1.99999994039535 vs 1.605723403591
Sn, n=26: 1.99999997019767 vs 1.60720269353183
Sn, n=27: 1.99999998509883 vs 1.60857443564431
Sn, n=28: 1.99999999254941 vs 1.60984994584839
Sn, n=29: 1.9999999962747 vs 1.61103900649049
Sn, n=30: 1.99999999813735 vs 1.6121501176016
Sn, n=31: 1.99999999906867 vs 1.61319070032792
Sn, n=32: 1.99999999953433 vs 1.61416726282792
Sn, n=33: 1.99999999976716 vs 1.61508553647347
Sn, n=34: 1.99999999988358 vs 1.61595058837658
Sn, n=35: 1.99999999994179 vs 1.6167669149072
Sn, n=36: 1.9999999999709 vs 1.61753851984547
Sn, n=37: 1.99999999998545 vs 1.61826898003539

I am trying to show that as n approaches infinity, Sn ours is always less than Sn theirs. And since we know Sn theirs approaches 2, Sn ours must be less than 2. But I am not sure how to prove it.
Blah! I don't know. I hope it's helpful even though I got stalled...

Answer 5

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Answer 6

Well, it is because it equals (pi^2)/6, which is less than 2.

Here's a proof (not trivial).
http://mathforum.org/library/drmath/view/56969.html

Done.

Answer 7

I would be very happy to do your homework = Slacker

sum (1 to infinity) of 1/(n^2) is less than 2
sum (1 to infinity) of 1/(n^2) is less than 2 and more than 1/(.8*n)(*) is the assumption of the divisional asset of .2N* and consequently N&*& is binary of ^(.0.133443443) which ecludes all explainations of why it's imparitive it is less then 2 and why you should do your own homework.

Answer 8

you will make money by learning this stuff, because it will get you good grades. good grades lead to a good gpa, hence a good job.

your retarded if you don't think learning dumb stuff for grades does not correlate with future earnings

Answer 9

You're wasting your time studying junk like that. It will never make you a dime..