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- Thread starter Lisa91
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$\displaystyle \sum_{n=1}^{\infty}(-i)^na_n=\sum_{n=1}^{\infty}(-1)^na_{2n}+i\sum_{n=1}^{\infty}(-1)^na_{2n-1}=a+bi$

Therefore:

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n}+\sum_{n=1}^{\infty}(-1)^na_{2n-1}=a+b$

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{n}=a+b$

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I guess the main idea is to show that the series [tex] \sum_{n=1}^{\infty}(-1)^na_{n} [/tex] is a part of it so it must be convergent but I don't really see it.

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$\displaystyle \sum_{n=1}^{\infty}(-i)^na_n=\sum_{n=1}^{\infty}(-1)^na_{2n}+i\sum_{n=1}^{\infty}(-1)^na_{2n-1}=a+bi$

Then we must have:

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n}=a$

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n-1}=b$

and so:

$\displaystyle \sum_{n=1}^{\infty}(-1)^na_{2n}+\sum_{n=1}^{\infty}(-1)^na_{2n-1}=\sum_{n=1}^{\infty}(-1)^na_{n}=a+b$

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[tex] \sum_{n=1}^{\infty}(-1)^na_{2n} [/tex]

[tex] -a_{1}+a_{3}-a_{5}+... [/tex]

[tex] \sum_{n=1}^{\infty}(-1)^na_{2n-1}[/tex]

[tex] -a_{2}+a_{4}-a_{6}+... [/tex]

So we get:

[tex] -a_{1}-a_{2}+a_{3}+a_{4}-a_{5}-a_{6}+... [/tex]

[tex] \sum_{n=1}^{\infty}(-1)^na_{n} [/tex] the series differs a liitle bit.

- Feb 13, 2012

- 1,704

According to the so called 'Abel's Test' if for a complex sequence $a_{n}$ is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$ and the series...Let's take [tex]\sum_{n=1}^{\infty} (-i)^{n} a_{n} [/tex], which is convergent , [tex] a_{n} > 0 [/tex]. What can we say about the convergence of this one: [tex]\sum_{n=1}^{\infty} (-1)^{n} a_{n}?[/tex] What can I do with it?

$\displaystyle f(z)= \sum_{n=0}^{\infty} a_{n}\ z^{n}$ (1)

... converges when $|z|<1$ and diverges when $|z|>1$ and the coeffcints $a_{n}$ are positive real numbers monotonically tending to 0, then f(z) converges everywhere on the unit circle except in z=1. Are You in such conditions?...

Kind regards

$\chi$ $\sigma$

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[tex] |z|= |0^{2}+(-1)^{2}| = |1| [/tex] so according to the rule we don't know whether it is convergent or not...According to the so called 'Abel's Test' if for a complex sequence $a_{n}$ is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$ and the series...

$\displaystyle f(z)= \sum_{n=0}^{\infty} a_{n}\ z^{n}$ (1)

... converges when $|z|<1$ and diverges when $|z|>1$ and the coeffcints $a_{n}$ are positive real numbers monotonically tending to 0, then f(z) converges everywhere on the unit circle except in z=1. Are You in such conditions?...

Kind regards

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

If You consider the series...[tex] |z|= |0^{2}+(-1)^{2}| = |1| [/tex] so according to the rule we don't know whether it is convergent or not...

$\displaystyle \sum_{n=0}^{\infty} a_{n}\ z^{n}$ (1)

... You know that it converges for $z=-i$. That implies that, because $(-i)^{n}$ doesn't tend to 0 with n, is $\displaystyle \lim_{n \rightarrow \infty} a_{n}=0$. You know also that for all n is $a_{n}>0$. Two of the requirement of the Abel criterion are satisfied, so that You can [preliminary...] conclude that the series converges on all the unit circle with the exception of z=1 [so that it converges for z=-1...] if for all n 'large enough' is $a_{n+1}<a_{n}$. Now we have to analyse if the last limitation can be in some way overcomed...

Kind regards

$\chi$ $\sigma$

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- #11

You're absolutely right; what I suggested doesn't work.

[tex] \sum_{n=1}^{\infty}(-1)^na_{2n} [/tex]

[tex] -a_{1}+a_{3}-a_{5}+... [/tex]

[tex] \sum_{n=1}^{\infty}(-1)^na_{2n-1}[/tex]

[tex] -a_{2}+a_{4}-a_{6}+... [/tex]

So we get:

[tex] -a_{1}-a_{2}+a_{3}+a_{4}-a_{5}-a_{6}+... [/tex]

[tex] \sum_{n=1}^{\infty}(-1)^na_{n} [/tex] the series differs a liitle bit.

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- Feb 7, 2012

- 2,752

Under those conditions, the series $\sum_{n=1}^{\infty} (-1)^{n} a_{n}$ need not converge. Suppose for example that $a_n = \begin{cases} 1/n^2 & (n \text{ odd}) \\ 1/n & (n \text{ even}) \end{cases}$. Then $$\sum_{n=1}^{\infty} (-i)^{n} a_{n} = \biggl(i\sum_1^\infty \frac{(-1)^n}{(2n-1)^2}\biggr) + \biggl(\sum_1^\infty \frac{(-1)^n}{2n}\biggr),$$ and both of the bracketed series converge. But $$\sum_{n=1}^{\infty} (-1)^{n} a_{n} = -1+\frac12 -\frac1{3^2} + \frac14 -\frac1{5^2} + \frac16 - \ldots\,.$$ In that series, the negative terms form a convergent series, but the positive terms form a divergent series, and so the whole series must diverge.Let's take [tex]\sum_{n=1}^{\infty} (-i)^{n} a_{n} [/tex], which is convergent , [tex] a_{n} > 0 [/tex]. What can we say about the convergence of this one: [tex]\sum_{n=1}^{\infty} (-1)^{n} a_{n}?[/tex] What can I do with it?