Question

Finding the sum of the series (Will post the problem below) (I thank you all in advance for contributing)

Answer 1

\[\sum_{3}^{\inf} \frac{ 1 }{ k(k+4) }\]

Answer 2

Im not quite understanding how to start a problem. Would I have to separate this by partial sums?

Answer 3

most likely it's a telescope series. So yes

Answer 4

Ok how do I do that?

Answer 5

Im stuck on that part

Answer 6

Have you already made it into partial fractions, show what you've gotten so far.

Answer 7

After you have the partial fractions, start writing out the first few sums of the series. You'll notice after a few terms that you will see the same terms repeat, only negative.

Answer 8

Is this correct?
\[(\frac{ 1 }{ k }-\frac{ 1 }{ k+4 })\]

Answer 9

that multiply by 1/4

Answer 10

Almost, except that they're both divided by 4.

Answer 11

So

\[(\frac{ 1 }{ 4k }-\frac{ 1 }{ 4k+16 })\]

Answer 12

Sure, and I'd probably just leave it outside rather than distributing it out like that to make it easier. Just multiply your final answer by 1/4.

Answer 13

yes, to see the pattern easier, i would suggest take out 1/4 and see the patter of
1/k - 1/(k+4)

Answer 14

Oh ok

Answer 15

\[\frac{1}{4}(\sum_{k=3}^{\infty} \frac{1}{k}-\sum_{k=3}^{\infty} \frac{1}{k+4})\] So here it is easier to see split up.

Notice that after 4 terms with the last sum, you'll be adding the same terms that were just subtracted 4 terms ago by the other term. So you can see how it simplifies to \[\frac{1}{4}\sum_{k=3}^{6} \frac{1}{k}\]

Which isn't an infinite sum and easy to add up on your own.

Answer 16

Actually writing out the terms rather than doing it all in summation is probably better for you to do so you understand what I'm saying. Write out the first 4 terms and from then on you'll see that the rest of the terms exactly cancel. =) Try it out and we'll help you finish it.

Answer 17

ok so the first 4 terms of the partial sum are

(4/21)+(1/8)+(4/45)+(1/15)

Answer 18

When they are added up