analysis
Name:
Anonymous
2008-11-03 13:35
is the infinite series (1/4 + (-1)^n)/n absolutely convergent?
Name:
Anonymous
2008-11-03 13:57
It looks like some kind of harmonic series to me, so I guess it diverges.
Name:
Anonymous
2008-11-03 14:01
No. Absolute value of terms is 5/4n or 3/4n. Taking the lower of the two gives sum from n = 1 to infinity of 3/4n, which is divergent.
Name:
Anonymous
2008-11-03 14:03
It's late, so I might have made an error somewhere.
\sum_{n=1}^{\infty} \frac{\frac{1}{4} + (-1)^n}{n}
= \sum_{n=1}^{\infty} \left(\frac{1}{4n} + \frac{(-1)^n}{n} \right)
= \frac{1}{4} \sum_{n=1}^{\infty} \left(\frac{1}{n} + \frac{(-1)^n}{n} \right)
= \frac{1}{4} \sum_{n=1}^{\infty} \frac{(-1)^n + 1}{n}
If n is equal =>
(-1)^n + 1 = 2.
If n is odd =>
(-1)^n + 1 = 0.
Therefore, we can write:
= \frac{1}{4} \sum_{n=1}^{\infty} \frac{2}{2n} = \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n}
Which is the harmonic series. The harmonic series is divergent.
Name:
Anonymous
2008-11-03 14:06
>>4
Yes, I made an error somewhere. Sorry, this is obviously wrong.
Name:
Anonymous
2008-11-03 14:07
>>5
You pulled out the 1/4 incorrectly after the second equal sign.
Name:
Anonymous
2008-11-03 21:45
>>4
What do you use to input math symbols?
I havn't found any good software....
Name:
Anonymous
2008-11-03 22:35
>>7
[math]\int^he_uses
\, tags
Name:
Anonymous
2008-11-04 0:11
\sum_{n=1}^{\infty} \frac{\frac{1}{4} + (-1)^n}{n}
= \sum_{n=1}^{\infty} \left(\frac{1}{4n} + \frac{(-1)^n}{n} \right)
= \frak{1}{4}\sum_{n=1}^\infty \frak{1}{n} + \sum_{n=1}^\infty \frak{(-1)^n}{n}
= \infty + 2
= \infty
looks like this works
Name:
Anonymous
2008-11-04 0:12
\sum_{n=1}^{\infty} \frac{\frac{1}{4} + (-1)^n}{n}
= \sum_{n=1}^{\infty} \left(\frac{1}{4n} + \frac{(-1)^n}{n} \right)
= \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n} + \sum_{n=1}^\infty \frac{(-1)^n}{n}
= \infty + 2
= \infty
woops, this should work now
Name:
Anonymous
2008-11-04 1:19
What's the absolute value of (1/4 + (-1)^n)/n ?
Name:
Anonymous
2008-11-04 3:58
thanks guys
Name:
Anonymous
2008-11-04 4:07
Is this infinite series convergent? ( 1/4 + 2/(n+1) )^n
Name:
Anonymous
2008-11-04 9:39
nevermind about this
>>13
Name:
Anonymous
2008-11-06 10:50
Is your dick convergent ?
Name:
Anonymous
2008-11-06 17:33
>>10
\sum^{\inf}_{1} \frac{(-1)^n}{n} = Log(2)
>>13
Obviously yes, for n > 1 its terms are bounded above by (11/12)^n and below by 0.
Name:
Anonymous
2008-11-06 20:15
>>9
doesn't know how to
/frak
Name:
Anonymous
2008-11-08 23:40
>>13
\lim_{n \rightarrow \infty} \left( \frac{1}{4} + \frac{2}{n+1} \right)^n \text{ is zero if } \lim_{n \rightarrow \infty}\left( \frac{1}{4} + \frac{2}{n+1} \right) < 1\\
\\
\lim_{n \rightarrow \infty}\left( \frac{n+1+8}{4(n+1)} \right) = \lim_{n \rightarrow \infty}\left( \frac{1}{4} \right)
So yes.
Name:
Anonymous
2008-11-08 23:44
\lim_{n \rightarrow \infty} \left( \frac{1}{4} + \frac{2}{n+1} \right)^n \text{ is zero if } \lim_{n \rightarrow \infty}\left( \frac{1}{4} + \frac{2}{n+1} \right) < 1\\
\\
\lim_{n \rightarrow \infty}\left( \frac{n+1+8}{4(n+1)} \right) = \lim_{n \rightarrow \infty}\left( \frac{1}{4} \right)
Name:
Anonymous
2008-11-08 23:44
\lim_{n \rightarrow \infty} \left( \frac{1}{4} + \frac{2}{n+1} \right)^n \text{ is zero if } \lim_{n \rightarrow \infty}\left( \frac{1}{4} + \frac{2}{n+1} \right) < 1
\lim_{n \rightarrow \infty}\left( \frac{n+1+8}{4(n+1)} \right) = \lim_{n \rightarrow \infty}\left( \frac{1}{4} \right)
Name:
Anonymous
2008-11-11 9:44
what language is that, latex? im not good with this
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