1. ## Series

Hi

I am find it hard to figure this series out?

Would anyone have any ideas?

1 +2/3+ 8/15 + 48/105 +384/945+ 3840/10395 + 46080/135135 + 645120/2027025 +10321920/34459425+...

2. Originally Posted by hebby
Hi

I am find it hard to figure this series out?

Would anyone have any ideas?

1 +2+ 8 + 48 +384+ 3840 + 46080 + 645120 +10321920+...
3 15 105 945 10395 135135 2027025 34459425
This is

$T_1=1$

$T_{n+1}=T_n(2n)$

3. sorry the numbers were incorrect

4. Originally Posted by hebby
Hi

I am find it hard to figure this series out?

Would anyone have any ideas?

1 +2/3+ 8/15 + 48/105 +384/945+ 3840/10395 + 46080/135135 + 645120/2027025 +10321920/34459425+...
This is $T_1=1$

$T_{n+1}=T_n\left(\frac{2n}{2n+1}\right)$

5. hi

then how would you write it in the sigma form?

I have to use the ratio test

6. Hi hebby,

there is no need to use the sigma format in this case.

$a_{n+1}=a_n\left(\frac{2n}{2n+1}\right)$

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

As n approaches infinity, the numerator and denominator become indistinguishable, hence the limit of this ratio as n approaches infinity is 1.
The ratio test is inconclusive for this series, it does not tell you if the series converges or diverges.

$\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\li m_{n\rightarrow\infty}|\frac{T_{n+1}}{T_n}|=1$

7. Hi

When I apply the Raabe Test,

Rn= lim (n(1-(an+1/an)), I get 1/2, does that mean it diverges?

8. the partial sum is $\sum^{n}_{k=0}\frac{4^{k}}{ (2k+1){2k \choose k} }=\frac{4^{n+1}}{ {2n+2 \choose n+1} }-1.$

9. Originally Posted by hebby
Hi

When I apply the Raabe Test,

Rn= lim (n(1-(an+1/an)), I get 1/2, does that mean it diverges?
Yes, hebby.

There is no question of conditional convergence, since it's not an alternating series.

10. Originally Posted by hebby
Hi

I am find it hard to figure this series out?

Would anyone have any ideas?

1 +2/3+ 8/15 + 48/105 +384/945+ 3840/10395 + 46080/135135 + 645120/2027025 +10321920/34459425+...
Show by induction that $T_n\geqslant\frac{1}{n}$