## A summation question

Hi

Given the sum

$$\sum _{p=0} ^{\infty} (-1)^p \frac{4p+1}{4^p}$$

I have tried something please tell if I'm on the right track

Looking at the alternating series test

(a) $$1/(4^{p+1}) < (1/(4^p))$$

(b) $$\mathop {\lim }\limits_{p \to \infty } b_p = \mathop {\lim }\limits_{p \to \infty } \frac{1}{{4^p }} = 0$$

Then according to the test this allows me to write $$\sum _{p = 0} ^{\infty} 4^{-p} = 4/3$$

Can anybody please verify if I'm heading in the right direction on this? Or am I totally wrong?

Sincerely Yours

Hummingbird

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 I agree that you've shown this series converges, but I don't see where you're getting that you can say it converges to $\frac{4}{3}$. The Alternating Series Test can show conditional convergence, but not a numerical value to the best of my knowledge.

Okay thanks I can see that now,

but what would be the next logical step to find the sum of this series? Should I use a specific test?

Sincerely Yours
Hummingbird25

p.s. Since it converges, then |1/(4^p)| < 1 ??

 Quote by Jameson I agree that you've shown this series converges, but I don't see where you're getting that you can say it converges to $\frac{4}{3}$. The Alternating Series Test can show conditional convergence, but not a numerical value to the best of my knowledge.

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## A summation question

 Quote by Hummingbird25 but what would be the next logical step to find the sum of this series? Should I use a specific test?
This is another of those that can be broken up with one of the summands being of the form $$px^p$$. Remember the method I suggested in your other thread ?

Hello and the other sum being

(-1)^p ?

Sincerely Hummingbird25

 Quote by Curious3141 This is another of those that can be broken up with one of the summands being of the form $$px^p$$. Remember the method I suggested in your other thread ?

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 Quote by Hummingbird25 Hello and the other sum being (-1)^p ? Sincerely Hummingbird25
No, the other summand is $$(-4)^{-p}$$.

That's just a geometric series.