HI there, just stuck on a question I can't figure out, would like some help.
The original question:
Determine whether the series converges (infinite series, from 1 to infinity)
(k+1)! / (k+1)^k
I know this can be rewritten as an infinite series from 0 to infinity of
k! / k^k
After that, I have no idea how to formally prove that this series converges. I know it of course it does, I just don't know how to prove it.
Help? =) Thanks!
The original question:
Determine whether the series converges (infinite series, from 1 to infinity)
(k+1)! / (k+1)^k
I know this can be rewritten as an infinite series from 0 to infinity of
k! / k^k
After that, I have no idea how to formally prove that this series converges. I know it of course it does, I just don't know how to prove it.
Help? =) Thanks!
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Well, it doesn't translate directly as the sum from 0 to ∞ of k!/k^k. (The power would have to be k + 1 in the denominator.) It is the same as
∞
Σ k (k!/k^k)
k=0
There is an extra factor of k. If you use the ratio test, calling the kth term a_k
a_(k+1)/a_k = [(k+1)/k] k^k/(k+1)^k = [(k+1)/k] (1 + 1/k)^(-k).
If you take the limit as k->∞, the factor (k+1)/k goes to 1. The remaining factor has limit
lim (1 + 1/k)^(-k) = 1/e.
k->∞
So taken together, you have
lim a_(k+1)/a_k = 1/e.
k->∞
Since this number is less than 1, the series converges by the ratio test.
∞
Σ k (k!/k^k)
k=0
There is an extra factor of k. If you use the ratio test, calling the kth term a_k
a_(k+1)/a_k = [(k+1)/k] k^k/(k+1)^k = [(k+1)/k] (1 + 1/k)^(-k).
If you take the limit as k->∞, the factor (k+1)/k goes to 1. The remaining factor has limit
lim (1 + 1/k)^(-k) = 1/e.
k->∞
So taken together, you have
lim a_(k+1)/a_k = 1/e.
k->∞
Since this number is less than 1, the series converges by the ratio test.
-
Σ (k+1)! / (k+1)^k
Apply the ratio test:
lim k→∞ | ((k+2)! / (k+2)^(k+1)) / ((k+1)! / (k+1)^k) |
lim k→∞ | [(k+2)! (k+1)^k] / [(k+1)! (k+2)^(k+1)] |
lim k→∞ | [(k+2)(k+1)k! (k+1)^k] / [(k+1)k!(k+2)(k+2)^k] |
(k+2), (k+1), and k! in the numerator/denominator cancel out.
lim k→∞ | (k+1)^k / (k+2)^k | = L
lim k→∞ ln( |(k+1)^k / (k+2)^k| ) = ln(L)
using log properties
lim k→∞ (k ln( (k+1)/(k+2) ) = ln(L)
lim k→∞ (k ln( (1+1/k)/(1+2/k) ) = ln(L)
lim k→∞ ( ln((1+1/k)/(1+2/k)) / (1/k) ) = 0/0
Apply L'Hopital's rule:
lim k→∞ [ ((k+2) - (k+1)) / (k+2)²) / (-1/k²) ]
lim k→∞ [ (1/(k+2)²) / (-1/k²) ]
lim k→∞ [ -k² / (k² + 4k + 4) ]
lim k→∞ [ -1 / (1 + 4/k + 4/k²) ] = -1 = ln(L)
ln(L) = -1
e^(-1) = L
1/e = L
With the ratio test, if L<1, the series converges absolutely; if L>1, the series diverges.
1/e < 1, so the series converges by the ratio test.
Apply the ratio test:
lim k→∞ | ((k+2)! / (k+2)^(k+1)) / ((k+1)! / (k+1)^k) |
lim k→∞ | [(k+2)! (k+1)^k] / [(k+1)! (k+2)^(k+1)] |
lim k→∞ | [(k+2)(k+1)k! (k+1)^k] / [(k+1)k!(k+2)(k+2)^k] |
(k+2), (k+1), and k! in the numerator/denominator cancel out.
lim k→∞ | (k+1)^k / (k+2)^k | = L
lim k→∞ ln( |(k+1)^k / (k+2)^k| ) = ln(L)
using log properties
lim k→∞ (k ln( (k+1)/(k+2) ) = ln(L)
lim k→∞ (k ln( (1+1/k)/(1+2/k) ) = ln(L)
lim k→∞ ( ln((1+1/k)/(1+2/k)) / (1/k) ) = 0/0
Apply L'Hopital's rule:
lim k→∞ [ ((k+2) - (k+1)) / (k+2)²) / (-1/k²) ]
lim k→∞ [ (1/(k+2)²) / (-1/k²) ]
lim k→∞ [ -k² / (k² + 4k + 4) ]
lim k→∞ [ -1 / (1 + 4/k + 4/k²) ] = -1 = ln(L)
ln(L) = -1
e^(-1) = L
1/e = L
With the ratio test, if L<1, the series converges absolutely; if L>1, the series diverges.
1/e < 1, so the series converges by the ratio test.