## anonymous one year ago Examine the following series for convergence:

1. anonymous

|dw:1442906412331:dw|

2. anonymous

Eugh sorry that looks terrible. Wanted to get something down real fast

3. anonymous

Sum of (a^k/k^p) where a> 0 and p>0

4. Empty

 $\sum_{k=1}^\infty \frac{a^k}{k^p}$ $\sum_{k=1}^\infty \frac{a^k}{k^p}$

5. anonymous

Right. Thank you. I should take the time to learn that since I'll probably be frequenting this site more often >.<

6. Empty

Hahaha it's fine I was just trying to help so you could see what it looks like and play around, I'm still a little distracted and so I just went ahead and put that, I'll help in a minute

7. anonymous

Alright then. Thank you. 1 other tiny technical thing, when I post a message there's like a ghost message that remains in the box and I have to refresh to be able to type again. Have you ever seen that?

8. Empty

Yeah the site itself here is pretty terrible unfortunately.

9. Empty

Ahhh alright, so I'm free now, admittedly it's been a while since I've done this so I'm not entirely sure but you might be able to confirm, this looks like something the ratio test will be able to show us convergence on, have you tried that yet?

10. anonymous

Oh wait, I did check that but I realize I made a dumb mistake. Let me double check some things

11. anonymous

So taking the k+1 over the kth term we have the limit as k goes to infinity of a(k^p) / (k+1)^p|dw:1442907554307:dw|

12. Empty

Yeah, I think it's safe to say that: $a\frac{k^p}{(k+1)^p} <a \frac{k^p}{k^p} = a$ So your limit will converge to a as you approach infinity. I'm not entirely sure if that means your series diverges or not, or if this just means $$a \le 1$$ I forget the specifics of this test.

13. Empty

For instance, it might be something slightly more nuanced than this, I forget as far as alternating series are concerned go, so it's possible that my $$\le$$ sign should be $$<$$ sign. I guess we really don't need to worry about a lower bound since we're already given that a and p are greater than 0 which is convenient.

14. anonymous

Okay so if the ratio is always less than a, then for all a values less than 1 it converges, no matter what p value we select right?

15. anonymous

Can we say anything about when a is greater than or equal to 1 from this?

16. Empty

We can say that it will definitely diverge for $$a \ge 1$$. Why? Because this is the ratio of successive terms, and if the ratio between successive terms is greater than 1 it means the terms are getting larger. If the ratio between successive terms is equal to 1, then it means that the terms are the same, which means it's not decreasing! Here are some examples to make it clearer: 1, 2, 4, 8, 16, ... The ratio between terms here is 2 (which is greater than 1), clear to see I think that this will diverge. 5, 5, 5, 5, 5, ... The ratio here between terms is 1, this clearly won't converge! 1, 1/2, 1/4, 1/8, 1/16, ... The ratio here between terms is 1/2, which is less than one, and adding all these terms together looks like it has the possibility of converging which is what we're looking for and this sort of thinking more generally explains why we require that: $a < 1$ Since this is the ratio between successive terms. :D

17. anonymous

Right, that makes sense, but we're also dividing by k^p right? So if p is very large couldn't that "cancel out" how fast a is growing?

18. anonymous

Plus from the ratio test we only have that it is less than. So if a were say, 5, a(k^p)/(k+1)^p could be .9 and the statement would still be true.

19. Empty

No, because we know p>0 that means no matter how big p is, the bigger it gets, the _smaller_ our terms are since it's in the denominator.

20. Empty

I think you're confusing the ratio test with the terms of the series. The ratio test compares two arbitrary successive terms of the series, it's kind of confusing so I'll try to figure out how to explain it better depending on your response to try to get at where you're having trouble.

21. anonymous

Well in either scenario, in the terms or in the ratio test, it seems like a large p could make it reasonable that it would converge. So in the terms, if the series were just a^k, we would know automatically that it converges for a<1 and diverges otherwise. and if it were just 1/k^p then we would know automatically that it converges if p >1. and diverges otherwise . So in the combined, its easy to know that it would definite converge if a<1 and p>1, and vice versa for diverge But when there's overlap between the converge part and divergent part, there could be leeway.

22. Empty

Here's another way I just found to show this that doesn't use the ratio test: First, look at the geometric series: $\sum_{k=1}^\infty a^k$ We might already know this converges for $$a<1$$ already. for k>0 and p>0 we know that $$\frac{1}{k^p}<1$$ If you multiply something by a value less than 1, it will make it smaller. We can multiply this by every term in the geometric series to get this: $\sum_{k=1}^\infty \frac{1}{k^p}a^k$ Which will also converge for a<1 as well.

23. anonymous

In terms of the ratio test we have a(k^p)/(k+1)^p < a. If a is, say, 6. Then changing the p value could vary the quantity anywhere from 0 to 5.9 or so, and if its 0-1 its convergent 1-5.9 divergent

24. Empty

Wait I think I see what you're saying, is it possible that for values like $$a \ge 1$$ is there some p we can choose so that it converges?

25. anonymous

Right

26. anonymous

27. Empty

I am not sure how to answer your question rigorously, but I think not since the part with a is exponential and the part with p is linear, so I don't think it will happen.

28. anonymous

Oh okay, yeah right, that makes sense.

29. Empty

Yeah, good question though.

30. Empty

It might be worth looking into more if you're curious to find a proof of what you're asking, not sure how I'd go about doing it though unfortunately. Maybe math stackexchange can help?

31. anonymous

So even for values like a =2 and p = 1,000,000,000, there's no chance that they'll approach 0? Sorry if this is super foundation stuff, just want to make sure.

32. Empty

Well, it looks like if you pick a=1 you'll get the harmonic series which diverges unless you pick p>1. So you can converge for this special case it seems. Beats me, this is weird. I think for a>1 though you'll come into some real problems because a will start to exponentially grow in power with each successive term while for any p you pick, you will always reach some threshold where you'll pass it on your way to infinity.

33. anonymous

That makes sense. Thank you.

34. anonymous

Alright, thank you for all your help. I know this was a huge time sink for you but I really do appreciate it.

35. Empty

Haha it's fine I mean it's interesting and fun don't worry about it. If I don't have time for this I won't log in :P

36. anonymous

That's fair, just know that you really helped me out a lot with this. Not just this problem but overall understanding.

37. Empty

Awesome :D Yeah please tag me if you want some more help, don't feel like you're a burden or anything like that I want to do calculus problems and stuff that's why I come on here and I enjoy helping!

38. anonymous

What you guys did is perfectly fine. The conditions are $\lim \left| \frac{ a_{k+1} }{ a_{k} } \right| < 1$ for convergence. Using the ratio test you end up with $\lim_{k \rightarrow \infty} a(\frac{ k }{ k+1 })^{p} = a$ Since p is fixed, we can just consider $$1^{p}$$ to be 1, So the convergence of the series ends up depending only upon a.