# Thread: series convergence or divergence

1. ## series convergence or divergence

Does this series diverge or converge:

1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...

2. Originally Posted by WartonMorton
Does this series diverge or converge:

1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
This is $\displaystyle \sum_{n=1}^\infty a_n$, where $\displaystyle a_1=1$, and $\displaystyle a_n=a_{n-1}\cdot\tfrac{n}{2n-1}$, and, yes, this series converges.
For consider that $\displaystyle \lim_{n\to\infty}\frac{n}{2n-1}=\frac{1}{2}<1$

3. Originally Posted by WartonMorton
Does this series diverge or converge:

1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
$\displaystyle a_{n+1}=a_n \frac{n+1}{2n+1}$

so:

$\displaystyle \frac{a_{n+1}}{a_n}=\frac{1}{2-\frac{1}{n+1}}$

CB

4. Originally Posted by WartonMorton
Does this series diverge or converge:

1 + [1*2]/[1*3] + [1*2*3]/[1*3*5] + [1*2*3*4]/[1*3*5*7} + ...
$\displaystyle 1 + \frac{1 \cdot 2}{1 \cdot 3} + \frac{1 \cdot 2 \cdot 3}{1 \cdot 3 \cdot 5} + \frac{1 \cdot 2 \cdot 3 \cdot 4}{1 \cdot 3 \cdot 5 \cdot 7} + \dots$

$\displaystyle = \sum_{k = 1}^{\infty}{\frac{k!}{\frac{(2k)!}{k!2^k}}}$

$\displaystyle = \sum_{k = 1}^{\infty}\frac{(k!)^22^k}{(2k)!}$

Using the ratio test:

$\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{[(n + 1)!]^22^{n + 1}}{[2(n + 1)]!}}{\frac{(n!)^22^n}{(2n)!}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{(2n)![(n + 1)!]^22^{n + 1}}{[2(n + 1)]!(n!)^22^n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{2(2n)!(n + 1)^2(n!)^22^n}{(2n + 2)(2n + 1)(2n)!(n!)^22^n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{2(n + 1)^2}{2(n + 1)(2n + 1)}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{n + 1}{2n + 1}\right|$

$\displaystyle = \lim_{n \to \infty}\left(\frac{n + 1}{2n + 1}\right)$

$\displaystyle = \lim_{n \to \infty}\frac{1}{2}$ by L'Hospital's Rule

$\displaystyle = \frac{1}{2}$

$\displaystyle < 1$.

So the series is convergent.

5. Originally Posted by Prove It
$\displaystyle 1 + \frac{1 \cdot 2}{1 \cdot 3} + \frac{1 \cdot 2 \cdot 3}{1 \cdot 3 \cdot 5} + \frac{1 \cdot 2 \cdot 3 \cdot 4}{1 \cdot 3 \cdot 5 \cdot 7} + \dots$

$\displaystyle = \sum_{k = 1}^{\infty}{\frac{k!}{\frac{(2k)!}{k!2^k}}}$

$\displaystyle = \sum_{k = 1}^{\infty}\frac{(k!)^22^k}{(2k)!}$

Using the ratio test:

$\displaystyle \lim_{n \to \infty}\left|\frac{t_{n + 1}}{t_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{[(n + 1)!]^22^{n + 1}}{[2(n + 1)]!}}{\frac{(n!)^22^n}{(2n)!}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{(2n)![(n + 1)!]^22^{n + 1}}{[2(n + 1)]!(n!)^22^n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{2(2n)!(n + 1)^2(n!)^22^n}{(2n + 2)(2n + 1)(2n)!(n!)^22^n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{2(n + 1)^2}{2(n + 1)(2n + 1)}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{n + 1}{2n + 1}\right|$

$\displaystyle = \lim_{n \to \infty}\left(\frac{n + 1}{2n + 1}\right)$

$\displaystyle = \lim_{n \to \infty}\frac{1}{2}$ by L'Hospital's Rule

$\displaystyle = \frac{1}{2}$

$\displaystyle < 1$.

So the series is convergent.
This looks like very hard work compared to the two previous posts, I expect that explains why it took 11 minutes longer to write.

CB