∑n=1^∞ (-1)^n (ln n)/((1.01)^n) ∑n=1^∞(n!)/(n^n) ∑n=1^∞((n!)^2)/((2n)!) (2)/(1!) - (2 · 4)/(2!) + … + (-1)^n+1(2

$\sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{(1.01)^n}$ \\ $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ \\ $\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}$ \\ $\frac{2}{1!} - \frac{2 \cdot 4}{2!} + \dots + (-1)^{n+1} \frac{2 \cdot 4 \dots (2n)}{n!} + \dots$

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Transcript

00:01 Let's determine whether this series converges or diverges.

00:05 By writing the sn, this is recalled the end partial sum.

00:15 In our case, since we're starting at 2, this would just be a2 all the way up to a .n.

00:24 So we'll write this thing as a telescoping sum, as they showed in the examples in the textbook.

00:29 And then if it's convergent, we'll actually go ahead and find the sum.

00:35 So here, before we do anything, regarding the sum, let's just look at the a -n here.

00:43 So we have two.

00:45 Let's go ahead and factor that denominator.

00:52 And then here we would have to do partial fractions.

01:00 So we would have to solve this equation here for a and b.

01:04 We can multiply both sides by the denominator on the left.

01:09 And then you get a -n -minus -1, b -n -plus -1.

01:15 So here you can solve this for a and b and you end up with negative 1 for a, 1 for b.

01:26 So i'll need more room here to evaluate the sum.

01:30 So i'll need to go to the next page here.

01:33 Our sum is from 2 to infinity.

01:38 And then we just used a partial fraction.

01:41 So we have 1 over n minus 1, minus 1 over n plus 1.

01:47 This is after the partial fraction decomp.

01:52 Then, and of course, this is our n value...

Question

$\sum_{n=1}^{\infty} (-1)^n \frac{\ln n}{(1.01)^n}$ \\ $\sum_{n=1}^{\infty} \frac{n!}{n^n}$ \\ $\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}$ \\ $\frac{2}{1!} - \frac{2 \cdot 4}{2!} + \dots + (-1)^{n+1} \frac{2 \cdot 4 \dots (2n)}{n!} + \dots$

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