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Sum from 1 to infinity of (1+n)/((n)2^n) ~ is this right?

  • Thread starter Lo.Lee.Ta.
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  • #1
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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
 
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Answers and Replies

  • #2
SammyS
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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)(2n+1}}{\frac{1+n}{(n)(2n)}}[/itex]

= [itex]\frac{n(n+2)}{2(n+1)2}[/itex]

= [itex]\frac{n2 + 2n}{2(n+1)2}[/itex]

ALSO, DOES ANYONE KNOW WHAT'S THE MATTER WITH MY LATEX SYMBOLS?!
Yes. You can't use the [ SUP] [\SUP] tags in LaTeX.
I'm trying to write the same thing as what's below:

[(1+n+1)/[(n+1)(2n+1)]]/[(1+n)/((n)(2n))]

= n(n+2)/(2(n+1)2)

= (n2 + 2n)/(2n2 + 2n + 1)
______________________________________________________________________________

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
Are you only checking for convergence, or do you need to evaluate the series?
 
  • #3
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Yay, I figured out the Latex.

And I guess I only have to check for convergence, but would I be able to find what it actually converges to?
 
  • #4
SammyS
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[itex]\displaystyle \frac{1+n}{n}=\frac{1}{n}+1[/itex]
 
  • #5
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Okay, I didn't distribute 2 to all the terms in the denominator!

I think it should really be:

[itex]\frac{n^{2} + 2n}{2n^{2} + 4n + 2}[/itex]

But because we have 2 as the highest power in both the denominator and numerator, can't we use the ratio of the coefficients and just say that the expression goes to 1/2?

In the ratio test, if our answer is less than 1, it converges. If it's greater than 1, it diverges.

So in this case, our expression converges.

...Also, can we figure out what the number is that it converges to? Is that number 1/2?

Thanks! :)
 
  • #6
33,075
4,776
Okay, I didn't distribute 2 to all the terms in the denominator!

I think it should really be:

[itex]\frac{n^{2} + 2n}{2n^{2} + 4n + 2}[/itex]

But because we have 2 as the highest power in both the denominator and numerator, can't we use the ratio of the coefficients and just say that the expression goes to 1/2?
Yes, the limit is 1/2. An easier way to see this is:
$$ \lim_{n \to \infty}\frac{n^2 + 2n}{2(n + 1)^2} = \lim_{n \to \infty}\frac{n^2(1 + 2/n)}{2n^2(1 + 2/n + 1/n^2 } = 1/2$$
In the ratio test, if our answer is less than 1, it converges. If it's greater than 1, it diverges.
You would be clearer if you didn't use the word "it." In the sentence above, "it" refers to two different things.

In the ratio test, if our answer is less than 1, [STRIKE]it[/STRIKE] the series converges. If [STRIKE]it's[/STRIKE] the answer is greater than 1, [STRIKE]it[/STRIKE] the series diverges.
So in this case, our expression converges.

...Also, can we figure out what the number is that it converges to? Is that number 1/2?

Thanks! :)
No, the fact that the ratio of successive terms has a limit of 1/2 tells us only that the series converges, but not what the sum of the series is.
 

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