Question

I'd like to figure out this infinite product.

Answer 1

\[\prod_{n=1}^{\infty} (1+\frac{ 1 }{ n })^n\]

Answer 2

I'm gonna say it's probably divergent but not sure. All I know is that it's larger than e by just looking at it.

Answer 3

are you sure?

Answer 4

About?

Answer 5

it being greater than e

Answer 6

\[e=(1+\frac{ 1 }{ \infty })(1+\frac{ 1 }{ \infty })(1+\frac{ 1 }{ \infty })*...\]\[this=(1+\frac{ 1 }{ 1})(1+\frac{ 1 }{2})(1+\frac{ 1 }{ 3 })*...\]

So both have the same number of terms, but the terms in this problem are all larger... except the "last" term... lol.

Answer 7

I believe it's divergent to. I was comparing in it to an exponential function like
(1 + a)^n. this function will grow if a > 0

so (1 + a)^1 > (1 + a)^2 > (1+a)^3 > .. > (1 + a)^n

Answer 8

Hmm I feel like that's sort of on the right path, but since e approaches a finite number maybe this does as well? The limit as this product approaches infinity is multiplication by 1.

Answer 9

http://www-math.ucdenver.edu/~spayne/prelim/inf.pdf

Answer 10

I'm not interested in reading that link.

Answer 11

Here's an idea, how about turning the infinite product into a sum? \[\exp[ \sum_{n=0}^{\infty} n*\ln(1+\frac{ 1 }{ n })]\]

So this seems to be a divergent sum since it fails the divergence test since the limit of the sum = 1.

Answer 12

I was thinking (1 + 1/n)^n = (1+n)^n / n^n,

(1 + n)^n > n^n and so,
(1 + n)^n / n^n > 1

we multiply a number that is bigger than 1 infinitely many times, so the product grows

Answer 13

Well it's settled, the divergence test already showed that, I guess I should close this now.