- #1
Lo.Lee.Ta.
- 217
- 0
1.
∞
Ʃ (1+n)/[(n)(2n)]
n=1
2. When I see that there is an n as an exponent, I think to do the ratio test.
___________________________________________________________________________
[itex]\frac{\frac{1+n+1}{(n+1)(2^{n+1})}}{\frac{1+n}{(n)(2^{n})}}[/itex]
= [itex]\frac{n(n+2)}{2(n+1)^{2}}[/itex]
= [itex]\frac{n^{2} + 2n}{2(n+1)^{2}}[/itex]
Would equal ∞/∞, so I think I'm supposed to do L'Hospital's rule...
→L'Hospital's→ [itex]\frac{2n^{2} + 2}{4n + 2}[/itex]
I think it would still equal ∞/∞, so would I do L'Hospital rule again?
→L'Hospital's→ [itex]\frac{4n + 2}{4}[/itex]
It no longer equals ∞/∞, so would we then have: n + 1/2 = ∞
This really doesn't seem right... :/
Please help!
Thanks so much! :D
∞
Ʃ (1+n)/[(n)(2n)]
n=1
2. When I see that there is an n as an exponent, I think to do the ratio test.
___________________________________________________________________________
[itex]\frac{\frac{1+n+1}{(n+1)(2^{n+1})}}{\frac{1+n}{(n)(2^{n})}}[/itex]
= [itex]\frac{n(n+2)}{2(n+1)^{2}}[/itex]
= [itex]\frac{n^{2} + 2n}{2(n+1)^{2}}[/itex]
Would equal ∞/∞, so I think I'm supposed to do L'Hospital's rule...
→L'Hospital's→ [itex]\frac{2n^{2} + 2}{4n + 2}[/itex]
I think it would still equal ∞/∞, so would I do L'Hospital rule again?
→L'Hospital's→ [itex]\frac{4n + 2}{4}[/itex]
It no longer equals ∞/∞, so would we then have: n + 1/2 = ∞
This really doesn't seem right... :/
Please help!
Thanks so much! :D
Last edited: