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Another tricky sumby twoflower
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#1
Dec404, 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. 


#2
Dec404, 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] 


#3
Dec404, 07:40 AM

P: 368




#4
Dec404, 07:58 AM

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P: 3,147

Another tricky sum
Does this help?
[tex]\lim_{n > \infty} \left(1  \frac {1}{n}\right)^n = \frac {1}{e}[/tex] 


#5
Dec404, 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...



#6
Dec404, 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... 


#7
Dec404, 02:02 PM

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P: 12,016

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}^{n1}(1\frac{1}{n^{2}})^{i}[/tex] Hence, your originial partial sum can be written as: [tex]\sum_{n=1}^{N}\sum_{i=0}^{n1}\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.. 


#8
Dec404, 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...



#9
Dec404, 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](11/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.



#10
Dec404, 04:58 PM

P: 368




#11
Dec404, 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 11/(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 1y/2[/tex] holds for (positive) values of y that are close to zero, which can be done using only derivatives. 


#12
Dec504, 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 ;). 


#13
Dec604, 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)^{i1}\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. 


#14
Dec604, 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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