## Quiz me on infinite series!

For the discussion of math. Duh.

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BirdKiller
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### Quiz me on infinite series!

To prepare for my infinite series final exam that consists almost purely of computational problems, I'm trying to solve 300 problems that consists of infinite sequence (does it diverge or converge, if latter, to what value?) and infinite series (diverge or converge?) by Sunday.

Problem is, I'm running out of problems, I've been solving problems from the book and the Internet and my sources of problems are starting to dwindle. Furthermore, most of them deal with simple tests like geometric, telescoping, ratio, root, and etc. It's quite rare to find problems that deal with alternate series, integrals, complex numbers, etc.

So just make-up any problems that deal with infinite sequence or series that should be difficult but something a freshmen can solve. I'll post my work and solution here too.

Also, any problems with computing the Taylor/power series of a function also works fine too.
Last edited by BirdKiller on Fri Jun 06, 2008 9:23 am UTC, edited 2 times in total.

SimonM
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### Re: Quiz me in infinite series!

I expect you've probably seen these, and they're not too hard for an undergrad either, but anyway

$\sum_{k=1}^{\infty} {2k \choose k} \frac{1}{5^k}$

$\sum_{k=1}^{\infty} \frac{5^nF_n}{7^n}$ where Fn is the nth Fib number
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jmorgan3
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### Re: Quiz me in infinite series!

Find the power series for xx.

Find the power series for [imath]\int x^x dx[/imath]

For what complex numbers s will $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ equal zero?

Note: These are very hard problems. Especially the last one.
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BirdKiller
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### Re: Quiz me in infinite series!

For some reason, my forum browser couldn't read the math code, so I tried to read it the best I could:

1. Sigma(n = infinity, k = 1): 2k/5k

By geometric test where |r| = 2/5 < 1, this series is convergent, the sum being 5/3

2. Sigma(n = infinity, k = 1): 7n/(5nFn) where Fn is the nth digit of the Fibonacci number.

From the root test, the series is divergent since |7 / (5Fn1/n)| = 7/5 > 1 as n approaches infinity

Last one I'm not sure about.

Thanks for the problems both of you. If I don't reply soon, that means I went to sleep, but I will work on them.

SimonM
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### Re: Quiz me in infinite series!

BirdKiller wrote:By geometric test where |r| = 2/5 < 1, this series is convergent, the sum being 5/3

The sum isn't right.

Also, rather than 5 and 7, try 3 and 5 (and you've got it upside down anyway)
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BirdKiller
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### Re: Quiz me on infinite series!

I'll be back within 6 hours or so...hopefully (sleep + class), again, I really appreciate the problems you two sent me.

As for the first one, I think it's because I mixed up 2k to 2k. Again, my browser can't interpret the code for some reason.

Token
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### Re: Quiz me in infinite series!

jmorgan3 wrote:Find the power series for xx.

Find the power series for [imath]\int x^x dx[/imath]

For what complex numbers s will $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ equal zero?

Note: These are very hard problems. Especially the last one.

The last one isn't hard at all. The answer is "none".
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Fafnir43
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### Re: Quiz me on infinite series!

Yup - the RIemann hypothesis concerns values of the zeta function with Re(s)<1, at which point that formula becomes meaningless and you have to resort to analytic continuation.

BirdKiller wrote:1. Sigma(n = infinity, k = 1): 2k/5k

By geometric test where |r| = 2/5 < 1, this series is convergent, the sum being 5/3

Yes and no - that's correct, but the first sum was actually Sigma(n=infinity, k=1): (2k choose k)/5^k, which is a bit harder.

BirdKiller wrote:2. Sigma(n = infinity, k = 1): 7n/(5nFn) where Fn is the nth digit of the Fibonacci number.

From the root test, the series is divergent since |7 / (5Fn1/n)| = 7/5 > 1 as n approaches infinity

Again, that should be Sigma(n = infinity, k = 1): 5nFn/(7n). And are you sure that the nth root of the Fibonacci numbers tends to 1 as n tends to infinity?
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jmorgan3
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### Re: Quiz me on infinite series!

Fafnir43 wrote:Yup - the Riemann hypothesis concerns values of the zeta function with Re(s)<1, at which point that formula becomes meaningless and you have to resort to analytic continuation.

You're right. Apparently I should start reading entire wikipedia articles instead of just skimming for equations.

To the OP: my third problem was a (failed) attempt to get you to work on one of math's most famous open problems. The other two, however, are legitimate but insanely complicated (algebra-wise) problems discussed in this thread.
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BirdKiller
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### Re: Quiz me on infinite series!

One go before I go back to classes:

1. Sigma(n = infinity, k = 1): (2k)/5k = a(k)

Using the Ratio test, the f(k) = a(k+1) / a(k) = (2k + 2) / (10k)
Let k approach infinity such that f(k) = 1/5 < 1 => the series is convergent.

Again, my browser can't interpret the math code, and so it'll be great if someone tell me how I can enable my browser to read so.

SimonM
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### Re: Quiz me on infinite series!

That term isn't 2k/5^k, it is the binomial coefficient 2k, k times 1/5^k
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Qoppa
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### Re: Quiz me on infinite series!

BirdKiller wrote:Again, my browser can't interpret the math code, and so it'll be great if someone tell me how I can enable my browser to read so.
Read the sticky at the top of the page.

Code: Select all

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Something Awesome
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### Re: Quiz me on infinite series!

Here's a fun one that shouldn't be beyond your scope, though it may seem complicated.

Prove that the perimeter of the Koch Snowflake is infinite but its area finite.

Hint: Find the sequence/series for each and prove they converge/diverge.

BirdKiller
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### Re: Quiz me on infinite series!

While I'm doing these nightmarish problems, would anyone try to prove that

2n/n! = 0 as n approaches infinity?

Intuitively and by brute calculations, it's obvious, but if there's something that I learned in this course, it's the fact that you shouldn't trust intuition too much. More specifically, I'm trying to find the upper bound formula of this equation such that I can apply the squeeze theorem.

Sorry I haven't posted up my answers, I've been chipping away at them time to time however.

NathanielJ
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### Re: Quiz me on infinite series!

BirdKiller wrote:While I'm doing these nightmarish problems, would anyone try to prove that

2n/n! = 0 as n approaches infinity?

Try to write out a bit more explicitly what that fraction is, and it should become obvious. A couple more spoilered hints if you need them:

Hint 1:
Spoiler:
There's n things being multiplied in the numerator, and n things being multiplied in the denominator.

Hint 2:
Spoiler:
Rewrite it as (2/1) * (2/2) * (2/3) * (2/4) * ... * (2/n)
It should be obvious now.
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jmorgan3
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### Re: Quiz me on infinite series!

BirdKiller wrote:While I'm doing these nightmarish problems, would anyone try to prove that

2n/n! = 0 as n approaches infinity?

Intuitively and by brute calculations, it's obvious, but if there's something that I learned in this course, it's the fact that you shouldn't trust intuition too much. More specifically, I'm trying to find the upper bound formula of this equation such that I can apply the squeeze theorem.

Sorry I haven't posted up my answers, I've been chipping away at them time to time however.

Divide the (n+1)th term by the nth term. It should be obvious after that.
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Frimble
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### Re: Quiz me on infinite series!

One of my favorites, not hard but interesting.

Determine if the series: sin θ + sin2 θ + sin3 θ... is convergent or divergent for the following values of θ. Where the series converges, find the value of the limit:

π/2

π/4

π

3π/2

Hence or otherwise find an expression for the limit valid for any real value of θ.
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