A summation question

Hummingbird25

Hi

Given the sum

[tex]\sum _{p=0} ^{\infty} (-1)^p \frac{4p+1}{4^p}[/tex]

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

Looking at the alternating series test

(a) [tex]1/(4^{p+1}) < (1/(4^p))[/tex]

(b) [tex]\mathop {\lim }\limits_{p \to \infty } b_p = \mathop {\lim }\limits_{p \to \infty } \frac{1}{{4^p }} = 0[/tex]


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

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 [itex]\frac{4}{3}[/itex]. 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 ??

Curious3141

This is another of those that can be broken up with one of the summands being of the form [tex]px^p[/tex]. Remember the method I suggested in your other thread ?

Hummingbird25

Hello and the other sum being

(-1)^p ?

Sincerely Hummingbird25

Curious3141

No, the other summand is [tex](-4)^{-p}[/tex].

That's just a geometric series.