Is the Derivative of a Sum always Convergent?

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Is the following correct?

[tex]\frac{d}{dt}\sum_{n=0}^{\infty}\frac{2^{n}t^{n}}{(n+1)!}=\sum_{n=0}^{\infty}\frac{d}{dt}\frac{2^{n}t^{n}}{(n+1)!}[/tex]
 
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Is the derivative of a sum the sum of its derivatives?

Yes, I believe so.
 
Yes. As long as the summation variable is different from the derifination variable.
 
As Mute said, we can differentiate (or integrate) an infinite sum "term by term" as long as the convergence is uniform. Fortunately, that is a power series and power series always converge uniformly inside their radius of convergence.

This particular example has infinite radius of convergence so it can be differentiated "term by term" for all x.
 
Dear friends,

What i understood from the previous threads is that if the summation converges summation and differentiation can be interchangeable. Here i assume that summation and differentiation variables are different.


But my doubts are
1. Is it an iff statement?. That is differentiation and summation are interchangeable iff the summation converges.

2.Here my summation is attachment1 . This diverges.
But i suspect that (due to some reasons) attachment2 do not diverge(It would be a function dependent on N after substituting value for x, which is a natural no >=2).

Any sensible suggestions would be really helpful :)
 
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bincy said:
Dear friends,

What i understood from the previous threads is that if the summation converges summation and differentiation can be interchangeable.
Do you? Several of the responses told you that is NOT true. I, for example, told you that the differerentiation and summation can be interchanged if the sum converges uniformly. That is a stronger requirement than just saying "converges".

Here i assume that summation and differentiation variables are different.But my doubts are
1. Is it an iff statement?. That is differentiation and summation are interchangeable iff the summation converges.
If and only if the sum converges uniformly

2.Here my summation is attachment1 . This diverges.
But i suspect that (due to some reasons) attachment2 do not diverge(It would be a function dependent on N after substituting value for x, which is a natural no >=2).

Any sensible suggestions would be really helpful :)
 
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An example of where differentiation can fail is
[tex]ln(1-x)=\sum_{n=1}^{\infty} \frac{x^n}{n}[/tex]

At x=-1 this sum converges (to ln(2)) but if we try to differentiate we get
[tex]\sum_{n=0}^{\infty} x^n[/tex]
and this sum diverges at x=-1, even though the power series is right differentiable at that point
 
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Good example, Office Shredder! As I said before, a power series converges uniformly, and so is differentiable, inside its radius of convergence. Here, -1 is one endpoint of the interval of convergence, not inside it.
 
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Thanks.

To Office_Shredder:Your example was really helpful. There a series was converging but its derivative is not. In my case i suspect derivative is converging but series is not.