Multiplying Series: sigma ((-1)^n)*((x - 4)^n)/(n5^n)

[mikexz]

I was wondering if it possible to combine or split two series that are multiplied together and both are convergent?

For example

sigma (-1)^n)*((x-4)^n)/(n5^n))

Find interval of convergence WITHOUT using ratio test.

are we allowed to separate this into 2 sums (since -1^n is convergent)? so it becomes

[sigma (-1)^n)] * [sigma ((x-4)^n)/(n5^n))]

I would think series that go to infinity is like we are dealing with limits as the function approaches infinity, the limit properties show that

so can I do the same ???

thanks

 

[daon]

Re: Multiplying Series

It's nowhere near as simple as that.

(-1)^n is not convergent anyway. Even if it was, your answer is no. Think of the distributive property.

 

[mikexz]

If i can't split it, doesn' that mean that the property of limits is FALSE???

Are there assumptions or restrictions to the properties of limits that I don't know about??


why would (-1)^n diverge??? the terms cancel out.

 

[daon]

If i can't split it, doesn' that mean that the property of limits is FALSE???

Are there assumptions or restrictions to the properties of limits that I don't know about??


why would (-1)^n diverge??? the terms cancel out.

\(\displaystyle \sum _{n=0}^{\infty} (-1)^n = 1 - 1 + 1 - 1 + 1 - ...\)

"Grouping" does not work here. One could easily obtain different values for the infinite sum by grouping tricks.

If \(\displaystyle S_n\) is the partial sum we get the sequence:

\(\displaystyle \{S_n\}_0^{\infty} = \{1,0,1,0,\,\, ...\}\).

This sequence obviously diverges.

And no to your other question. The algebra would of course not work.

 

[mikexz]

if an was convergent, would that limit be true and would that allow me to split the series?

 

[daon]

if an was convergent, would that limit be true and would that allow me to split the series?

No. Look at:

\(\displaystyle \frac{6}{5} = \sum _{n=0}^{\infty} \frac{1}{6^n} = \sum _{n=0}^{\infty} \frac{1}{2^n}\frac{1}{3^n} \neq \sum _{n=0}^{\infty} \frac{1}{2^n} \cdot \sum_{n=0}^{\infty}\frac{1}{3^n} = 2\frac{3}{2}=3\)