how can I show that the sum of the sequence from 1 to infinity of (k/(k+1))^(k^2) converges or diverges?

Answer 1

I wonder if this is the exact problem since this is extremely convergent.

I used the root test

show there exists C and N s.t.
for all k > N |a_k|^(1/k) < C < 1

yielding:
|a_k|^(1/k) = (k/(k+1))^k < .5 for all K >= 1

Answer 2

You could use L'Hopital's rule to show that the limit of the sequence is zero, ie the last term is zero. That means the sum is finite.

Let y = [k/(k+1)]^(k^2)
ln(y) = k^2 * [lnk - ln(k+1)]
= [lnk - ln(k+1)] / ]k^-2]

Applying the rule:
limit of lny = -k^3 / 2 * [1/k - 1/(k+1)] = -infinity
Thus y --> 0 as k --> inf

*EDIT*
If you just take a large value of k and plug it into your calculator, you'll see that the sequence does indeed converge to 0 not 1.
k/(k+1) is always slightly less than 1, so raising a fraction to the infinite power gives 0 not 1.

Answer 3

Since the terms are positive, we can use the root test, because convergence is then equivalent to absolute convergence. If a_k = (k/(k+1))^(k^2), then (a_k)^(1/k) = (k/(k+1))^k = (1 - 1/(k+1))^k = [(1 - 1/(k +1))^(k+1)](k/(k+1). As k --> oo, (1 - 1/(k +1))^(k+1) --> 1/e and k/k+1 --> 1, so that (a_k)^(1/k) --> 1/e < 1. So, the series Sum (k/(k+1))^(k^2) converges.

Answer 4

Probably the easiest way (for this problem) is to show that the limit of the terms is a non-zero value. Since k/(k+1) -> 1 as k -> ∞, the individual terms approach 1 asymtotically,and the sequence has no upper bound and diverges.

HTH

Doug

Answer 5

a_k= [1-1/(k+1)) ]^k^2= e^(k^2*ln (1-1/(k+1))
compare it with e^-k = (1/e)^k geometric series with r=1/e
lim a_k/e^-k = e^lim k^2 *ln(1-1/(k+1))+k
ln(1-1/(k+1))= -1/(k+1)-1/2*1/(k+1)^2---- ( Mc Laurin)
so
k^2 ln(1-1/(k+1) +k = -k^2/(k+1)+k -1/2k^2/(k+1)^2 ---
=k/(k+1) -1/2k^2/(k+1)^2----- = 1/2
so the limit=e^1/2 and by the comparison test both series are of the same class.
As (1/e)^k converges so does the given series

Answer 6

lim [k-> +oo] [(k+1 - 1)/(k+1)]^(k^2) =
= lim [[1-1/(k+1)]^-(k+1)]^[-k^2/(k+1) ] =
limit e^-[k^2/(k+1)= lim e^-k = 0

So, you can apply any of the other criteria to classify it, as the previous answerers did.

Ana

Answer 7

you ought to use L'Hopital's Rule on it. As written in, this is a nil * infinity type (ok->infinity, e^(-ok)->0). you may desire to rewrite it as a ratio. So write it as ok/e^ok, then you definately could be conscious L'Hopital.