1. ## Series Convergences

Its a normal summation for K=0 and goes to infinite for$\displaystyle [(-1)^k]/(2k+1)$

how do I show this converges... I know it does but I forgot how, using limit comparison, comparison, divergence, ratio, root, abs. ratio, abs. root, or AST.

thanks again

Its a normal summation for K=0 and goes to infinite for$\displaystyle [(-1)^k]/(2k+1)$

how do I show this converges... I know it does but I forgot how, using limit comparison, comparison, divergence, ratio, root, abs. ratio, abs. root, or AST.

thanks again
Use the Alternatig Series Test using the fact that if $\displaystyle a_n=\frac{1}{2n+1}$ then $\displaystyle a_{n+1}<a_n$ and $\displaystyle \lim_{n\to\infty}a_n=0$

Note: $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$

3. so Im digging through my backpack and I find a paper explaining the AST for oscillating harmonics lol

thanks though..

Now tell me this... is a multiple of a harmonic only when its 1/k multiplied by a number > 1 ?

so like 1/2k this would not be a multiple of a harmonic?

because by AST, we also take the lim as x approaches infinte... 1/2k+1 compares to 1/k .. 1/2k+1 < 1/k aww which proves nothing.. darn it but the limit comp. text proves its 1/2 . Thus the two act similiar so it diverges.. or did I mess that up somewhere?

so Im digging through my backpack and I find a paper explaining the AST for oscillating harmonics lol
That is always how it goes

so Im digging through my backpack and I find a paper explaining the AST for oscillating harmonics lol

thanks though.. you truly are a math stud.. (no homo)

Now tell me this... is a multiple of a harmonic only when its 1/k multiplied by a number > 1 ?

so like 1/2k this would not be a multiple of a harmonic?

because by AST, we also take the lim as x approaches infinte... 1/2k+1 compares to 1/k .. 1/2k+1 < 1/k aww which proves nothing.. dammit but the limit comp. text proves its 1/2 . Thus the two act similiar so it diverges.. or did I mess that up somewhere?
No if you have $\displaystyle \sum \frac{a}{n}$ where $\displaystyle a$ is any real constant then your series converge. But I think you are confusing the AST, it states roughly: If you have a series $\displaystyle \sum (-1)^na_n$ and $\displaystyle a_{n+1}\leqslant a_{n}$ and $\displaystyle \lim a_n=0$ then the series converges.