00:01
In the first part of the question, we will solve our question as given here as n is equal to summation of n is equal to 1 to infinity for factorial n by n to the power of n.
00:18
So here we will use using ratio test, our ratio will be as limit n tends to infinity for a n to plus 1 divided by 8 n so this is as limit n tends to infinity n plus 1 factorial divided by n plus 1 to the power of n plus 1 multiplied by factorial n to the power of n divided by factorial n and now limit n tends to infinity n plus 1 n plus 1 n plus 1 multiplied by n to the power of n and denominator will be multiplied by n plus 1 to the power of n.
01:25
And now it is limit n tends to infinity for 1 by 1 plus 1 upon n to the whole power n.
01:38
Since we know that since 1 plus 1 upon n to the power of n is equal to e.
01:45
So we can write here r is equal to 1 upon a and then the series is convergent.
02:01
So here we have our solution for first part of the question.
02:05
This is the answer for a part.
02:08
Now let's see our second part of the question.
02:12
For part b, we will again take our sum of n terms which is given a summation of n is equal to 1 to infinity.
02:23
2 to the power of n, e to the power of minus n.
02:26
So again we will use our using ratio test.
02:34
Our ratio r will be as limit n tends to infinity for a n plus 1 by a n.
02:44
And now r values as limit n tends to infinity.
02:50
2 to the power of n plus 1 multiplied by e to the power of minus n plus 1.
02:58
Divided by 2 to the power of n e to the power of minus n...
