@TedShifrin Recitations or sectons is fine. I try to teach in these, so I don't like the term problem sessions. If a student wants to talk to me about problems, I am always available.
So, @Null, if we do $1-1+1-1+1-1+1-\dots$, the even partial sums are $0$ and the odd partial sums are all $1$. Is this not an example? The series does not converge.
@TedShifrin but I just want to be sure, $\sum_{k=1}^{2n+1} (-1)^k$ is convergent, if the top value is even? But isn't convergence defined this way: for any epsilon, we can choose an integer, such that after that, the sum is smaller then L+$\epsilon$?
$\|x+h\|^2 - \|x\|^2 = 2h \cdot x + h \cdot h$. Dividing out by magnitude of $h$, we get $2h \cdot x / \|h\|$, which is well-defined in the limit. Thus one concludes that $\|x\|^2$ is differentiable, and so taking square roots we're good away from zero.
Apparently that's what he did but his work looked way more complicated.
but if I have proven, that the limit of an odd(lets call this integer c) partial sum is something. And $s_{c-2}<s_{c-1}<s_c$ can be shown, have we proven that ??? lost my red thread haha
I think that will converge @Null. The convergence of odd partial sums $S_{2k + 1}$ together with the partial sums increasing would imply partial sums bounded above I believe
So monotone convergence comes in, so the partial sums converge.
In general, if $S_{2k+1}$ converges and $S_k$ also converges, then the limits will be the same
But in general, convergence of odd-indexed terms doesn't imply convergence of the whole thing.
haha, that feels dirty @TedShifrin. I was excited about 20k, but it turns out it's 25k I am excited about. I really want access to the Google analytics
ok: alternating series test for $\sum(-1)^k\frac{1}{\sqrt{k}}=\sum(-1)^ka_k$. The limit of $a_n$ is 0, check. Then we have to check if |$a_{n+1}|\leq |a_n|$. Simple stuff gives us that. So it converges by the alternating series test. done?
can some one help me with a derivative im confused with - im trying to find the time when velocity of my object is zero. but from my understanding there are 2 times when that is the case. that or i am misunderstanding how to work it out. i got my derivative to be v = 3t^2 - 6t - 9 , v = 0 when t is -1 or 3 seconds right? but the question is worded in a way that i only need to find a single moment in time which is confusing me
The classic example where you end up with a situation with two solutions is someone running at a constant speed trying to catch up with an accelerating bus.
If that speed is high enough, then you'll run right past the bus; eventually, the bus will pass again.
But if all you want is the time to catch the bus, then that second solution is irrelevant.
I have a very quick question about the definition of periodic functions. My textbook says that the period of any function, $T$ must always be greater than zero while other sources that I have found say that it is sufficient that $T \ne 0$. The latter seems to be correct but I wanted to make sure; so which is it?
