Solving Summation Question: Alternating Series Test

[Hummingbird25]

HELP: 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

 

[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.

 

[Hummingbird25]

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 ??

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.

 

[Curious3141]

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 ?

 

[Hummingbird25]

Hello and the other sum being

(-1)^p ?

Sincerely Hummingbird25

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 ?

 

[Curious3141]

Hello and the other sum being

(-1)^p ?

Sincerely Hummingbird25

No, the other summand is (-4)^{-p}.

That's just a geometric series.