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estro
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I have hard time with latex so I attached all relevant info in that picture file.
Thanks you in advance!
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System said:Do not make it complicated
take the limit(as n goes to infinity) of both sides of the inequality in the question (the inequality which has the nth rooth of the absolute value of an)
the left side will be simply the limit which is in the root test and the right side will be 1
so the limit of the nth root of the absolute value of an is less than 1
so the series converges by the nth root test.
System said:No.
Take the limit of both sides as I said, the limit of the right side will be 1
so the limit of the right side (which the nth root of the absolute value of an) is LESS THAN 1 ---> The series converges by the root test.
System said:Sir,
The inequality itself
take the limit of both sides
it will something like
the lim of the left side LESS THAN the lim of the right side
System said:your website said: if L < 1 then the series absolutely convergent (hence convergent).
Its agree with my conclusion :)
System said:...
then you could consider the inequality starts from n=2, and this will be remove the "EQUAL" sign from the inequality
...
estro said:What?! (we are talking about LIMIT)
System said:If f(n)<g(n) for n>a
then:
f(n)<g(n) for n>a+1
I'm not sure but I think it is not. Again root test can't be used here.(despite the resemblance)Red_CCF said:I'm not sure if this is right but what I did was:
let bn = 1 - 1/n^a, for 0 < a < 1, and an = bn ^ n; by virtue of the root test, this converges since the limit is between but not equal to 0 and 1.
Dick said:Good for you for resisting all of the bad root test advice. You could try to apply an integral test to exp(-n^(1-a)). It gives you an incomplete gamma function, if you know how to deal with that. I mostly know this because I typed it into Maxima. It diverges as a->1, of course, but seems to be finite on a in (0,1).
estro said:Thank you for your response.
I'm not too familiar with the gamma function and how to deal with it,
nevertheless I tried applying integral test to exp(-n^(1-a)), ( too much work=) and not sure how to integrate it anyway).
Do you think there is no better way, and I should make it with the integral test?
Dick said:The integral test isn't that bad, especially if you remember that you don't need to find an exact value for the integral. You just need to show the integral converges. Change variables to u=n^(1-a). Can you show the integral is u^K*exp(-u) for some constant K?
estro said:Thank you for your guidance.
Here is my raw attempt, I'm on right way? (My integration is a little nasty...)
View attachment 26252
Dick said:Double check that. If you do it right the new integral should be t^(a/(1-a))*exp(-t). That means the power of t is positive, so t^(a/(1-a)) never becomes less than 1. You want to show this converges by showing exp(-t) goes to zero fast enough. exp(t) grows faster than any polynomial in t as t->infinity, right?
estro said:Oh, I blundered with my integration.
But now how can I show that t^(a/1-a)/e^t goes to 0 fast enough?
Dick said:Call K=(a/(1-a)). Then for any N you can find t large enough so that exp(t)>t^N. Pick N=K+2 for example.
estro said:But 1<x for every x>1 and still integral_(1/x)dx from 2 to infinity is not convergent.
I have trouble showing integral_(t^K/e^t)dt is convergent.
Dick said:I'm not suggesting you show t^K/e^t is less than 1. I'm suggesting you show it's less than 1/t^2 for large t.
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