Yes I apologize. When writing up the question, I got excited that this lemma would help me prove a bigger problem I had. Indeed, ##G/N## is a group since ##N\vartriangleright G##, but of course ##G/N## itself is not a subgroup of ##G##, which renders the lemma useless in my case, though, I do...
Homework Statement
Let ##G## be a group of order ##n## where ##n## is an odd squarefree prime (that is, ##n=p_1p_2\cdots p_r## where ##p_i## is an odd prime that appears only once, each ##p_i## distinct). Let ##N## be normal in ##G##. If I have that ##|G/N|=p_j## for some prime in the prime...
Well, ##\log(1+a_n) < a_n## so it will converge when ##a_n## does?
Also, if it does converge then wouldn't
##\sum_{n=1}^{N} \log(1+a_n) = \log(1+a_1)+\log(1+a_2)+\ldots+\log(1+a_n) = \log(\prod_{n=1}^{N}(1+a_n))## imply that the product would converge?
Homework Statement
a_n is a sequence of positive numbers. Prove that \prod_{n=1}^{\infty} (1+a_n) converges if and only if \sum_{n=1}^{\infty} a_n converges.
Homework Equations
The Attempt at a Solution
I first tried writing out a partial product: \prod_{n=1}^{N} (1+a_n) =...
Oops
I guess in my "excitement" I neglected the powers.
So we ACTUALLY obtain \frac{1}{n[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3}]}, and now n^{2/3} < (n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3} so that \frac{1}{n[(n+1)^{2/3}+n^{1/3}(n+1)^{1/3} + n^{2/3}]} < \frac{1}{n^{\frac{5}{3}}}, which...
Homework Statement
Does \sum_{n=1}^{\infty}a_n where a_n = \frac{(n+1)^{1/3}-n^{1/3}}{n} converge or diverge?
Homework Equations
The Attempt at a Solution
The ratio test is inconclusive, as is the root test. The limit is equal to 0, but that doesn't say much. I've tried to find...