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November 23rd, 2014, 10:40 AM  #1 
Newbie Joined: Jan 2012 Posts: 1 Thanks: 0  Need Help Understanding Infinite Series
I would appreciate it, if someone would explain to me how it can be that an infinite series can possibly converge to a finite number. If infinity means "never ending," then an infinite number of terms in a series should add to an infinite sum, not a finite number; however, according to my calculus book, an infinite series such as 1 + 1/2^2 + 1/3^2 + 1/4^2 + ... converges to (pi^2)/6. This doesn't make sense, to me; how can you finally arrive at a definite, specific sum, if you are forever adding to the partial sums? Yet the socalled "harmonic" series 1/1 + 1/2 + 1/3 + 1/4 + ...diverges to infinity. This does make sense to me, since you are continually and forever adding more & more to the partial sums, so of course the sum will be infinite; nobrainer, it seems. In both cases, we are forever adding on more and more, and in both cases, the terms are heading toward zero as the series progresses, so what can be the difference? Why should one end, while the other explodes into infinity? It would be most helpful, if any attempt to answer this seeming conundrum is given in plainEnglish, rather than in some complex, mathematical expression, since my math skill level isn't that great, as you can tell. It has been my experience that, if someone can't put an idea into simple terms, then they probably don't really understand it that well, themselves. Obviously, that idea has its limits, since some things, such as quantum physics, are, by nature, rather hard to water down, but I don't think that holds true in this case. Last edited by rayroshi; November 23rd, 2014 at 10:44 AM. 
November 23rd, 2014, 11:09 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
$$ \lim_{n\to\infty}\sum_{k=1}^nf(k)=c $$ mean that for any $\varepsilon>0$ that you choose, you can find some number N such that $$ \leftc\sum_{k=1}^nf(k)\right<\varepsilon $$ for any n >= N. With your example, let's take $\varepsilon=0.001$. You can compute that $$ \frac{\pi^2}{6}\sum{k=1}^{1000}\frac{1}{k^2}\approx0.0009995002 $$ and so you can view the convergence of the sum as the statement that no matter how large you choose n > 1000, the sum will be within 0.001 of $\pi^2/6.$ When you gain mathematical maturity you will see why we call this behavior "convergence", but that's not so important here. Quote:
$$ \sum_{k=1}^n\frac{1}{k}>x $$ which shows that the harmonic series is unbounded above. For example, you pick x = 1000 and I pick, oh, 10^435. It's possible to show (though it's too big to compute directly!) that $$ \sum_{k=1}^{10^{435}}\frac1k\approx1002.2017 $$ and you could do similarly for any choice of $x$. A series is unbounded above (and hence diverges) if, whenever we play this game, I can always win. If you can win (by choosing some huge number that I can't surpass by any choice of $n$) then the number is bounded above. If no terms are negative, then this means that the sum converges to some finite value. (Obviously negative terms can cause problems, like $\sum_{k=1}^\infty1$; in fact they can cause more subtle problems too with what is called conditional convergence but that's getting more complicated and you asked for easy.) *** In summary: both convergence and divergence can be viewed as twoplayer games which involve only finite numbers. Don't let the squiggle $\infty$ scare you away, because all it's doing is telling you to play one of these two games.  

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