Real Analysis - Sample Midterm Help

[steelphantom]

My professor has posted a sample midterm on her web site, but although she promised to post the solutions as well, she hasn't yet and I don't really expect her to at this point since the midterm is tomorrow. I have a few questions about some of the problems on the midterm. The sample exam can be found here: http://www.math.psu.edu/li/math312/hw.html

2(c): $$\sum$$ (2 / (3 - (-1)ⁿ))ⁿ
This series diverges, but how would I go about proving it? The first thing I thought of was the root test, but the limit of the series does not exist. What other way could I try to do this?

3: Consider a nonnegative series $$\sum$$a_n. If there exists an M such that A_N = $$\sum_{n=1}^N$$a_n <= M for all N, then $$\sum$$a_n is convergent.
This doesn't say to explicitly prove this, but I assume that's what I need to do. Isn't this true from the definition of convergence of infinite series?

For problems 6-8, it says "Find the limit of xxx." I can just stare at them and figure out what the limit is, but that's obviously not what my professor wants me to do. I guess the only thing I could do is say what the limit is and then prove it using the definition of a limit.

I know I should have asked my prof personally about this stuff, but I couldn't make her office hours yesterday, and the next office hours are Thursday, which will do me no good. As always, thanks for any help.

 

[quasar987]

For 2(c), notice that 3-(-1)^n is always bigger than 2, so (2 / (3 - (-1)n))n is always less than 2/2^n

3: It indeed follows from the definition. You are told that the partial sums are bounded by M... and recall that if a sequence is monotone and bounded, then it converges. :-o

for 6-8, this is exactly what you must do... figure out what your intuition tells you about the limit to those things, then prove using the dfn of limit that your intuition is right

 

[steelphantom]

Thanks for your help, quasar! But for 2(C), since I'm trying to prove the series diverges, don't I want to find a series that is smaller than it, rather than larger than it, if I want to try to use the comparison test?

 

[NateTG]

Thanks for your help, quasar! But for 2(C), since I'm trying to prove the series diverges, don't I want to find a series that is smaller than it, rather than larger than it, if I want to try to use the comparison test?

In order for a series to converge, the limit of the terms must be zero. This series fails that test since every other term is 1.

 

[steelphantom]

In order for a series to converge, the limit of the terms must be zero. This series fails that test since every other term is 1.

Well that was easy! Thanks for pointing that out.

 

[symbolipoint]

I'm tagging this question on the end, even though I really should create a separate topic instead:

Would restudyng the first 2 years-worth of undergraduate Calculus several times be helpful for later studying Real Analysis?

 

[quasar987]

Sorry to have mislead you for 2(c), I didn't realize that that n exponent was affecting the whole argument and not just the denominator.

For your question, I would be inclined to answer "no". Well yes, it would certainly be helpful, how can it hurt? But there are better ways to use your time.