00:01
In the part a of the problem, the given series is summation of n equals to 1 up to infinity minus 1 raised to the power of n which is divided by 5 multiplied with n.
00:17
So we can write this series as what? as 1 divided by 5 multiplied with n equals to 1 up to infinity minus 1 raised to the index of n divided by n.
00:32
So what we will do, we will apply the alternating series test.
00:42
So according to the alternating series test, we write that what an should be what? be positive.
00:55
And what and the limit of n tends to infinity over a n should be what? should be equals to 0.
01:08
So now therefore in this particular series, we see that what that limit n tends to infinity over 1 divided by n equals to what? equals to 0.
01:23
That is what? that is it converges.
01:27
Alright, so we see that there can be two conditions over here.
01:33
That is, it can be either absolutely convergent or it can be what? it can be conditional convergence.
01:41
Alright, it can be conditional convergence.
01:44
Alright, so now we see that summation of a n is what is absolutely convergent.
01:51
It is absolutely convergent when summation of modulus of a n what? is convergent.
02:02
Alright, so therefore we write that a n is conditionally.
02:08
We write that summation of a n is conditionally convergent or we just simply write conditional when what? when summation of what? of modulus of a n is what? is divergent.
02:26
Is divergent and the other condition is that and the summation of simple a n is what? it is convergent...