Help needed. (series)?

Question

Determine whether the series is convergent or divergent. If it is convergent, find its sum
a)(1/e^(2n)) n=1 and goes to infinity
b)Find the values of x for which the series converges. Use this to obtain the sum of the series for those values of x
2^n((x+1)^n) n=0 and goes to infinity

Answer 1

a) S= (1/e^2)^1 + (1/e^2)^2 + (1/e^2)^3 + ...
Se^2 = 1 + (1/e^2)^1 + (1/e^2)^2 + (1/e^2)^3 + ... = 1+S
S = 1/(e^2-1)

b) S = ∑ [2(x+1)]^n ; n=0 to ∞

This will converge when |a(n+1)/a(n)| = |2(x+1)|< 1
or x>-3/2 and x<-1/2

The sum is 1/(1-2(x+1)) = -1/(2x+1)

Lemma:
If it converges, S = ∑ a^n = 1/(1-a) for n = 0 to ∞

Proof:
S = 1 + a + a^2 + a^3 + ...
aS = a + a^2 + a^3 + ... = S-1
S(a-1) = -1 or S = 1/(1-a)
In this case a=2(x+1) so S = 1/(1-2(x+1)) = -1/(2x+1)

NOTE: 0^0 is usually considered an indeterminate form. If you allow 0^0=1, then S=1 when x=-1 (S = ∑0^n). However, many consider 0^0 as undefined so in this case x cannot equal -1.

NOTE TO PASCAL, edited: Thanks for double checking your analysis. We all make mistakes (and I am certainly not immune!). I believe it is incumbent upon all of us to check other answers, especially in the math section, where even a typo can totally change an answer.

Answer 2

Let's see: for your first one, the series converges, since the ratio of the nth term to the (n-1)th term is less than one. To find the value, let S=(n=1, ∞)∑1/((e²)^n). Then e²S=1+(n=1, ∞)∑1/((e²)^n), so by substituting, e²S=1+S. Subtract S to get (e²-1)S=1, and finally divide to get S=1/(e²-1)≈0.1565

For the second, note first that we can simplify 2^n((x+1)^n) by writing (2x+2)^n. This series will converge for precisely those values of x where -1<(2x+2)<1 (note that this inequality must be strict). Thus we get -3<2x<-1, and thus -3/2.

Edit 2: Ack! You're right Scott, I screwed this up. How embarrassing. I should have multiplied by the reciprocal of the n=1 term, which is what I did the first time. The correct derivation using this method would be S=(n=0, ∞)∑(2x+2)^n → S/(2x+2)=1/(2x+2) + (n=0, ∞)∑(2x+2)^n → S/(2x+2)=1/(2x+2)+S → S=1+(2x+2)S → (2x+1)S=-1 → S=-1/(2x+1), which is the same as your answer. I apologize for criticizing your solution without first thoroughly checking my work.

Answer 3

a) use the test of divergent, because as n goeas to infinite f(n) = 0 .

sorry i gotta go to watch simsons! i will come back and will explain it in detail.