Testing Series Convergence/Divergence: ∑ (-1)^(n)/ (1+1/n)^(n^2)

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Test the following Series for Convergence (absolute or conditional) or divergence

∑ (-1)^(n)/ (1+1/n)^(n^2)

I know we solve it with the root test but i reached at a point where i don't know how to cancel it out

----

lim(n--> infinity)= 1/(1+1/n+1)^(n+1^2) *[ (1+1/n)^(n^2)]/1

What do i do after this?
 
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<br /> \ldots=\frac{(1+\frac{1}{n})^{n^2}}{(1+\frac{1}{n+1})^{n^2+2n+1}}=<br /> (1+\frac{1}{n+1})^{-2n-1}\left[\frac{(n+1)^2}{n(n+1)}\right]^{n^2}<br /> \sim e^{-2}\left(\left[1+\frac{1}{n(n+1)}\right]^{n^2+n}\left)^\frac{n^2}{n^2+n}<br /> \sim e^{-1}<br />
 
Simkate said:
Test the following Series for Convergence (absolute or conditional) or divergence

∑ (-1)^(n)/ (1+1/n)^(n^2)

I know we solve it with the root test but i reached at a point where i don't know how to cancel it out

----

lim(n--> infinity)= 1/(1+1/n+1)^(n+1^2) *[ (1+1/n)^(n^2)]/1

What do i do after this?

You are using the ratio test, not the root test. The problem is a lot easier if you actually use the root test.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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