- #1
fauboca
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[tex]\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n[/tex]
z in C.
This only converges for [itex]z>\frac{-1}{2}[/itex], correct?
Thanks.
z in C.
This only converges for [itex]z>\frac{-1}{2}[/itex], correct?
Thanks.
vela said:No. For one thing, z is complex, so it doesn't make sense to say z > -1/2. In any case, show us how you reached this conclusion.
vela said:You need to explain what you did in more detail.
jbunniii said:Your inequality is true if and only if
[tex]\left|\frac{z}{z+1}\right|^2 < 1[/tex]
You can rearrange this and simplify to find a simple inequality for z.
Another way: your inequality is true if and only if
|z| < |z + 1|
What does this mean geometrically?
fauboca said:Why did you square it?
jbunniii said:Because you can then simplify [itex]|z + 1|^2[/itex]. If you're not sure how, try writing it as z + 1 times its complex conjugate.
fauboca said:I don't understand what you mean.
z + 1 times it conjugate is
[tex]z^2 - 1[/tex]
jbunniii said:No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.
fauboca said:So am I supposed to get [itex]\bar{z}> -1[/itex]??
jbunniii said:No. That doesn't even make any sense, as [itex]\bar{z}[/itex] is a complex number.
[tex]|z+1|^2 = (z+1)(\overline{z + 1}) = (z + 1)(\bar{z} + 1) = z\bar{z} + z + \bar{z} + 1 = |z|^2 + z + \bar{z} + 1[/tex]
So what does your inequality become?
Hint: you can simplify [itex]z + \bar{z}[/itex].
fauboca said:I defined [itex]z = x + yi[/itex] and then did what you said. But we had [itex]|z| < |z+1|[/itex] That is how I obtained my solution.
The convergence of a series is determined by whether the sequence of partial sums approaches a finite limit as the number of terms approaches infinity. In other words, if the value of the series approaches a specific number as more and more terms are added, then the series is said to be convergent.
The convergence of a series is directly related to the convergence of its individual terms. If the terms of a series do not approach zero, then the series cannot converge. However, the converse is not always true, as a series with terms that approach zero can still diverge.
To determine the convergence of this series, we can use the ratio test. This test involves taking the limit as n approaches infinity of the absolute value of the ratio between the (n+1)th term and the nth term. If this limit is less than 1, then the series is convergent. If it is greater than 1, the series is divergent. If it is equal to 1, the test is inconclusive.
No, the series will only converge for certain values of z. In this case, the series will converge for all values of z that satisfy $\left|\frac{z}{z+1}\right|<1$. This means that the series will converge for all complex numbers z with a magnitude less than 1.
Yes, there are several other tests that can be used, such as the integral test, comparison test, and root test. These tests can be particularly useful for determining the convergence of more complex series that cannot be easily evaluated using the ratio test.