Determine whether the series converges or diverges?

Question

Sumnation of (1+1/n)^2*e^-n where n=1 to infinite

Answer 1

I'mnot sure if you mean [(1+1/n)^2]*e^(-n) (which is what follows from what you wrote) or [(1+1/n)]^(2*e^-n).

In the first case, for every n, [(1+1/n)^2]*e^(-n) = e^(-n) + 2 e^(-n)/n + e^(-n)/(n^2). Since Sum e^(-n) is a convergente geometric series, it follows by comparisson that Sum 2 e^(-n)/n and Sum e^(-n)/(n^2) also converges. So, the series converges.

In the second case, which, I guess, is what you mean, for every n, [(1+1/n)]^(2*e^-n) = [(1+ 1/n)^n]^(2e^(-n)/n). Since (1 + 1/n)^n increases to e and 2e^(-n)/n --> 0, it follows that [(1+1/n)]^(2*e^-n) --.> e^0 =1. Since the sequence does not converge to 0, it follows the series diverges.

Answer 2

a) word that the words of the series are bounded by using (5/6)^n>(5^n)/(a million+6^n) because of the fact the denominator is greater effective in the left series, it converges by using the assessment try. b) As n-->oo, the words tend in direction of (n^3)/sqrt(n^8) (n^3)/(n^4) a million/n this is the harmonic series which diverges. So the unique series diverges by using the assessment try. c) very such as "a)" d) word that lim{n-->oo} a_n=10/4=5/2. which isn't 0, this implies the series ought to diverge by using the divergence try.

Answer 3

It's bounded above by 2e^(-n). And that converges.

So yes. :)