Another tricky sum


by twoflower
Tags: tricky
twoflower
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#1
Dec4-04, 04:36 AM
P: 368
Hi,

I really don't know which trick should I use to solve convergence of this sum:

[tex]
\sum_{n=1}^{\infty} a_{n}
[/tex]

where

[tex]
a_{n} = \left( 1 - \frac{1}{n^2} \right)^{n} - 1
[/tex]

Neither binomial development nor writing first few terms helped me, and because all terms of this sum are negative, I can't use many of theorems...

It seems it could have something to do with e, but I can't see the adjustment to get it exactly...

Thank you for any suggestion.
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Tide
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#2
Dec4-04, 06:54 AM
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Does this help?

[tex]1 - \frac {1}{n^2} = \left( 1 - \frac {1}{n}\right) \left( 1 + \frac {1}{n}\right)[/tex]
twoflower
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#3
Dec4-04, 07:40 AM
P: 368
Quote Quote by Tide
Does this help?

[tex]1 - \frac {1}{n^2} = \left( 1 - \frac {1}{n}\right) \left( 1 + \frac {1}{n}\right)[/tex]
Not much

Tide
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#4
Dec4-04, 07:58 AM
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Another tricky sum


Does this help?

[tex]\lim_{n -> \infty} \left(1 - \frac {1}{n}\right)^n = \frac {1}{e}[/tex]
twoflower
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#5
Dec4-04, 08:08 AM
P: 368
Yes I noticed it, but the only thing I was able to use it for was the prove of neccessary convergence condition. Anyway, I can't realize how to use this information in proving of convergence itself...
twoflower
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#6
Dec4-04, 08:09 AM
P: 368
I tried

[tex]
\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}} < 1
[/tex]

But it doesn't seem so...
arildno
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#7
Dec4-04, 02:02 PM
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Damn!
This one seems really difficult, and I have only one lame suggestion:
[tex](1-\frac{1}{n^{2}})^{n}-1^{n}=-\frac{1}{n^{2}}\sum_{i=0}^{n-1}(1-\frac{1}{n^{2}})^{i}[/tex]
Hence, your originial partial sum can be written as:
[tex]-\sum_{n=1}^{N}\sum_{i=0}^{n-1}\frac{1}{n^{2}}(1-\frac{1}{n^{2}})^{i}[/tex]

Possibly, this double sum may be rewritten in a clever manner as a Cauchy product, but I'm a bit skeptical..
twoflower
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#8
Dec4-04, 02:32 PM
P: 368
I took a look at what Cauchy product is (at wolfram.com) and I think that however it may lead to solving this problem, we're intended to solve it otherwise, because Cauchy product was not ever mentioned in the lecture...
shmoe
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#9
Dec4-04, 04:41 PM
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Is this a series from a practice final again? This is another question that I want to use Taylor series, or some bounds derived from series. You really need to know something about how fast [tex](1-1/n^2)^n[/tex] is approaching 1, and unless you've been given some inequalities to that effect (or some bounds like [tex]e^x\geq 1+x[/tex]) I don't see how you can do this.
twoflower
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#10
Dec4-04, 04:58 PM
P: 368
Quote Quote by shmoe
Is this a series from a practice final again? This is another question that I want to use Taylor series, or some bounds derived from series. You really need to know something about how fast [tex](1-1/n^2)^n[/tex] is approaching 1, and unless you've been given some inequalities to that effect (or some bounds like [tex]e^x\geq 1+x[/tex]) I don't see how you can do this.
No, this is just a problem from previous year winter semester from mathematical analysis test for informatics. It's definitely possible that it should be solved using Talyor's polynom, because according to sylabus we should learn it before the test comes...but now we're only at derivatives.
shmoe
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#11
Dec4-04, 05:30 PM
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I can't think of another way. The lower bound [tex]e^x\geq 1+x[/tex] (true for all x) in my last post is easy enough to prove just knowing derivatives, but I also want to use the upper bound [tex]e^{-1/x}\leq 1-1/(2x)[/tex] (for x>1) that won't be so easy to prove*.

If you assume these bounds (though you should be able to prove the first), you should be able to solve it. Try writing

[tex]\left( 1 - \frac{1}{n^2} \right)^{n}-1=e^{n\log\left( 1 - \frac{1}{n^2} \right)}-1[/tex]

And converting [tex]e^x\geq 1+x[/tex] to a bound for log, [tex]x\geq \log(1+x)[/tex]. Use this bound on the exponent. Then use the harder to prove upper bound.


edit*-on furthur thought, this won't be so difficult. You'll only need it to hold for large x. This is equivalent to proving [tex]e^{-y}\leq 1-y/2[/tex] holds for (positive) values of y that are close to zero, which can be done using only derivatives.
Tsss
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#12
Dec5-04, 09:43 AM
P: 5
[tex]n\ln(1 - \frac{1}{n^2})\sim-\frac{n}{n^2}=-\frac{1}{n}\rightarrow 0[/tex]

So, [tex](1 - \frac{1}{n^2})^{n}-1=e^{n\ln(1 - \frac{1}{n^2} )}-1\sim-\frac{1}{n}[/tex]

Now, you can conclude ;).
arildno
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#13
Dec6-04, 07:31 AM
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The easiest way to do this, is by making the binomial coeffecient expansion (I'll just change the sign):
[tex]1-(1-\frac{1}{n^{2}})^{n}=\sum_{i=1}^{n}(-1)^{i-1}\binom{n}{i}\frac{1}{n^{2i}}[/tex]
Now, you should be able to show that:
[tex]\binom{n}{i}\frac{1}{n^{2i}}\leq\frac{1}{i!n^{i}}[/tex]

Hence, each coefficient [tex]a_{n}[/tex] is an alternating (finite) series with terms of decreasing magnitude, hence:
[tex]a_{n}\geq\frac{1}{n}-\frac{1}{2n^{2}}\geq\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}[/tex]

Hence, your series diverges faster than one half of the harmonic series.
shmoe
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#14
Dec6-04, 10:18 AM
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That was nice and simple arildno! I gave up without too much thought on the binomial expansion and headed for a less direct route to get the same result (half the harmonic is what my string of inequalities will also yield).


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