find s(x)=1/x+4/x^2+9/x^3+16/x^4+... (x>1).

[yma16]

Someone told me to take derivative on both sides to get s(x) but I do not know how to do it. I figured out 1/x+2/x^2+3/x^3+4/x^4+...=x/(x-1)^2, (x>1) by breaking up every terms.

 

[sinx]

Someone told me to take derivative on both sides to get s(x) but I do not know how to do it. I figured out 1/x+2/x^2+3/x^3+4/x^4+...=x/(x-1)^2, (x>1) by breaking up every terms.

the general expression is;
s(x)=n²/xⁿ where n goes from 1 up.
is this what you are looking for?

 

[yma16]

yes, s(x)=the sum of all these terms.

some one wrote the answer for me but no details.

 

[sinx]

some one wrote the answer for me but no details.

I just did it by observation.
s(x)=1/x+4/x²+9/x³+16/x⁴+.....
i.e. I 'observed' the numerator is the square of the power of x (in the denominator).
1=1²,4=2², 9=3², etc.
therefore; s(x)=n²/xⁿ

 

[Steven G]

Someone told me to take derivative on both sides to get s(x) but I do not know how to do it. I figured out 1/x+2/x^2+3/x^3+4/x^4+...=x/(x-1)^2, (x>1) by breaking up every terms.

I am a bit confused as you said that you want to take the derivative of s(x) to get s(x). No no, if you take the derivative of s(x) you get s'(x), NOT s(x). Besides, even if took the derivative of s(x) to get s(x) this is a waste of time. Why? Because in order to take the derivative of s(x) you NEED s(x). Since you have s(x) you are done. s(x) = 1/x+2/x^2+3/x^3+4/x^4+.... So can you please explain exactly what you want?

 

[yma16]

what i want is the finite form of s(x)

for example 1/x+2/x^2+3/x^3+4/x^4+...=x/(x-1)^2. (x>1)

for example x/2+x/2^2+x/2^3+x/2^4+...=x

 

[yma16]

Finally, I understand the hint.

I am a bit confused as you said that you want to take the derivative of s(x) to get s(x). No no, if you take the derivative of s(x) you get s'(x), NOT s(x). Besides, even if took the derivative of s(x) to get s(x) this is a waste of time. Why? Because in order to take the derivative of s(x) you NEED s(x). Since you have s(x) you are done. s(x) = 1/x+2/x^2+3/x^3+4/x^4+.... So can you please explain exactly what you want?

In the result 1/x+2/x^2+3/x^3+4/x^4+....=x/(x-1)^2, its term is i/x^i. Take derivative of the term, it becomes -i^2/(x*x^i), denote it as (1). The right side's derivative is -(x+1)/(x-1)^3, denote it as (2).

If we move -1/x from each (1) to (2). Then (2) becomes x(x+1)/(x-1)^3. (1) becomes i^2/(x^i), which is the same as the term of s(x). Therefore, S(x)=x(x+1)/(x-1)^3, (x>1).

 

[Dr.Peterson]

In the result 1/x+2/x^2+3/x^3+4/x^4+....=x/(x-1)^2, its term is i/x^i. Take derivative of the term, it becomes -i^2/(x*x^i), denote it as (1). The right side's derivative is -(x+1)/(x-1)^3, denote it as (2).

If we move -1/x from each (1) to (2). Then (2) becomes x(x+1)/(x-1)^3. (1) becomes i^2/(x^i), which is the same as the term of s(x). Therefore, S(x)=x(x+1)/(x-1)^3, (x>1).

I think you confused us by not saying what you were told to take the derivative of! You seemed to be mentioning the simpler sum in passing, as a previous problem you had solved, not as what the first sentence referred to.

This time you have not said clearly what you mean by "move -1/x". I think you mean to multiply the entire left side by -x.

Clear communication is important both in asking others for help, and in seeing what you yourself are doing.

 

[tkhunny]

In the limit, you can also use older ideas.

1) 1/x + 4/x^2 + 9/x^3 + 16/x^4... = S
2) 1 + 4/x + 9/x^2 + 16/x^3 .. = Sx

Subtracting 2) from 1)

3) -1 - 3/x - 5/x^2 - 7/x^3 - ... = S(1-x)
4) -x - 3 - 5x - 7/x^2 - ... = Sx(1-x)

Subtracting 4) from 3)

5) x + 2 + 2/x + 2/x^2 + 2/x^3 + ... = S(1-x)^2
6) x + 2(1 + 1/x + 1/x^2 + 1/x^3 + ... ) = S(1-x)^2

Are we getting anywhere? Anyway, it's a good exercise in paying attention. There IS a Geometric Series in there, somewhere!

 

[JeffM]

Someone told me to take derivative on both sides to get s(x) but I do not know how to do it. I figured out 1/x+2/x^2+3/x^3+4/x^4+...=x/(x-1)^2, (x>1) by breaking up every terms.

This question is a mess.

Do you mean

\(\displaystyle \text {Given } \displaystyle n \in \mathbb Z^+, \ x > 1, \text { and } s_n(x) = \sum_{j=1}^n \dfrac{j}{x^j},\ \text {find } s_n'(x).\)

Or do you mean

\(\displaystyle \text {Given } \displaystyle n \in \mathbb Z^+ \text { and } x > 1, \text { simplify, if possible, } s_n(x) = \sum_{j=1}^n \dfrac{j}{x^j}.\)

Or do you mean something else altogether?

 

[yma16]

5) will lead to the same result

In the limit, you can also use older ideas.

1) 1/x + 4/x^2 + 9/x^3 + 16/x^4... = S
2) 1 + 4/x + 9/x^2 + 16/x^3 .. = Sx

Subtracting 2) from 1)

3) -1 - 3/x - 5/x^2 - 7/x^3 - ... = S(1-x)
4) -x - 3 - 5x - 7/x^2 - ... = Sx(1-x)

Subtracting 4) from 3)

5) x + 2 + 2/x + 2/x^2 + 2/x^3 + ... = S(1-x)^2
6) x + 2(1 + 1/x + 1/x^2 + 1/x^3 + ... ) = S(1-x)^2

Are we getting anywhere? Anyway, it's a good exercise in paying attention. There IS a Geometric Series in there, somewhere!

and it is more direct. Thanks.