- #1

- 864

- 17

It's just that last step I'm not getting, so you have:

[1 / K

^{z}] - [1 / (2K)

^{z}]

= [ (2K)

^{z}- K

^{z}] / [(2K)

^{z}* K

^{z}]

= [ (2)

^{z}- 1 ] / [(2K)

^{z}*]

Then what?

Thanks

- #1

- 864

- 17

It's just that last step I'm not getting, so you have:

[1 / K

= [ (2K)

= [ (2)

Then what?

Thanks

- #2

Mark44

Mentor

- 34,317

- 5,969

Expand the first series, and you get ##\frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + ... + \frac{1}{n^k} ... ##Hi

View attachment 80246

It's just that last step I'm not getting, so you have:

[1 / K^{z}] - [1 / (2K)^{z}]

= [ (2K)^{z}- K^{z}] / [(2K)^{z}* K^{z}]

= [ (2)^{z}- 1 ] / [(2K)^{z}*]

Then what?

Thanks

Expand the second series, and you get ##\frac{1}{2^k} + \frac{1}{4^k} + ... + \frac{1}{(2n)^k} ... ##

If you subtract the second series from the first, term by term, what do you get?

- #3

- 864

- 17

Sorry, I'm not sure, I'm too rusty.Expand the first series, and you get ##\frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + ... + \frac{1}{n^k} ... ##

Expand the second series, and you get ##\frac{1}{2^k} + \frac{1}{4^k} + ... + \frac{1}{(2n)^k} ... ##

If you subtract the second series from the first, term by term, what do you get?

- #4

Mark44

Mentor

- 34,317

- 5,969

Can you see that every term in the second series is also present in the first series?

- #5

- 864

- 17

Ah, I've been trying to use factorisation and cancellation.Can you see that every term in the second series is also present in the first series?

So when you minus S1 - S2 it leaves only 1/1

but how does that explain the final line of working they got for the sum?

- #6

Mark44

Mentor

- 34,317

- 5,969

No, that's not right. You still end up with a series with an infinite number of terms. Every term in the 2nd series is in the first series, but not every term in the 1st series is in the second. Expand both series like I did above, but with more terms, and then do the subtraction.Ah, I've been trying to use factorisation and cancellation.

So when you minus S1 - S2 it leaves only 1/1^{k}

tim9000 said:but how does that explain the final line of working they got for the sum?

- #7

mathman

Science Advisor

- 7,877

- 453

The second series consists of the reciprocals of powers of even integers. The first series consists of the reciprocals of all integers. Subtract the second from the first leaves the reciprocals of the powers of all odd integers.Sorry, I'm not sure, I'm too rusty.

- #8

- 864

- 17

No, that's not right. You still end up with a series with an infinite number of terms. Every term in the 2nd series is in the first series, but not every term in the 1st series is in the second. Expand both series like I did above, but with more terms, and then do the subtraction.

Woops, yes I see what you were saying. Aaah ok, no wonder I couldn't rearrange it, I was contemplating partial fractions etc.The second series consists of the reciprocals of powers of even integers. The first series consists of the reciprocals of all integers. Subtract the second from the first leaves the reciprocals of the powers of all odd integers.

Thanks

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