Correct. The first test for divergence should always be to see if the terms approach 0 as n increases. If they don't, then all other tests will be inconclusive or tedious to prove that it diverges.
actually you can show this using the root test. if the sequence converges to 1, but strictly from above, the series diverges en.wikipedia.org/wiki/Root_test
The fact that it grows bigger, even if it's a tiny bit tells us it's divergent. Let's say y = (1+1/x)^x. Then at x=5 for example, you get a value let's call it z . now because it's growing it will have a bigger value than z+z+z+z+z+z and so on. z+z+z+z and so on is divergent so it's divergent for sure
According to the fact n->inf (1+a/n)^(bn) = e^ab. Here a=-1 and b=1 (plug those in you get (1+-1/n)^n, therefore that equals to e^-1 = 1/e != 0, so it diverges by the test for divergence.
Teach I didn't get why we didn't use integral test 'cause if we say y=(1+1/n)^n and take ln of both sides we get -3n^2 and that is going to infinity and we proove that it diverges
I'm one of those who try to solve the question before playing the video and I just noticed that (1+1/n)^n is greater than 1^n, which is 1 since n > 0. I think it's called the comparaison theorem, but it's a fact (I think more widespread in math than other degrees) that any series S greater than a series T that goes to +infinity is divergent (smaller if T goes to -infinity). In this case, S is the series from the problem and T is the series 1+1+1+1+1... which is also known as the limit when n goes to infinity of n itself.
While technically correct, this is nonsense - this is clearly already divergent if you remove the +1/n part. It makes more sense to make this argument with (1-1/n)^n. Then the summands converge to 1/e.
For me it's obvious that it diverges even without the TFD. Like I'm only 14, I barely know how does it work, yet I still can conclude right away that it diverges
The terms you sum don't even tend to zero they're greater than 1. Even if it were to the power n^2. The end.
Correct. The first test for divergence should always be to see if the terms approach 0 as n increases. If they don't, then all other tests will be inconclusive or tedious to prove that it diverges.
I love how you have so many questions that appear on my home work!
actually you can show this using the root test. if the sequence converges to 1, but strictly from above, the series diverges en.wikipedia.org/wiki/Root_test
The fact that it grows bigger, even if it's a tiny bit tells us it's divergent. Let's say y = (1+1/x)^x. Then at x=5 for example, you get a value let's call it z . now because it's growing it will have a bigger value than z+z+z+z+z+z and so on. z+z+z+z and so on is divergent so it's divergent for sure
At 0:45 I knew the root test would't work because you said TRY the root test.
This is why you use the TFD first!
Thank you very much for the explanation, greetings from Argentina
what happens if we multiply a^n to the (1+(1/n))^n where a>0
I think we can also use binomial expansion to prove this
No it just makes things more complicated
What about (1 - 1/n)^n ?
I changed the plus to a minus. Does it also diverge by TFD?
According to the fact n->inf (1+a/n)^(bn) = e^ab. Here a=-1 and b=1 (plug those in you get (1+-1/n)^n, therefore that equals to e^-1 = 1/e != 0, so it diverges by the test for divergence.
Teach I didn't get why we didn't use integral test 'cause if we say y=(1+1/n)^n and take ln of both sides we get -3n^2 and that is going to infinity and we proove that it diverges
I'm one of those who try to solve the question before playing the video and I just noticed that (1+1/n)^n is greater than 1^n, which is 1 since n > 0. I think it's called the comparaison theorem, but it's a fact (I think more widespread in math than other degrees) that any series S greater than a series T that goes to +infinity is divergent (smaller if T goes to -infinity).
In this case, S is the series from the problem and T is the series 1+1+1+1+1... which is also known as the limit when n goes to infinity of n itself.
what if there is (1-1/n)^n^2?
Thank you so much, this really helped me!
While technically correct, this is nonsense - this is clearly already divergent if you remove the +1/n part. It makes more sense to make this argument with (1-1/n)^n. Then the summands converge to 1/e.
Awesome! Love your explanations!
Thank you very much sir ❤❤❤❤❤
Great Work Keep it up!!!
So basically the TFD is the same as the nth term test?
What if you take that and multiply by 1/5^n?
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It converges to a real number e.
The sum diverges
(1+¹/n)ⁿ as n approaches infinity equals e. Since e≠0, the series diverges.
For me it's obvious that it diverges even without the TFD. Like I'm only 14, I barely know how does it work, yet I still can conclude right away that it diverges
Ognjen Kovačević
TFD is just a formal way to say it diverges
Ok then.
I hope you accept my apology.
Incredible
Master
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