General Formula for Finding Sum of Complex Sequence | Homework Help"

[vishnu manoj]

Homework Statement

what is the general formula for the sequence (1/1*3+1/3*6+1/6*10+1/10*15...)

Homework Equations

i used the equation n/mn+1 but am not able to use it for this sequence

The Attempt at a Solution

I found the sequence of the denominators which is (1/2)n^2+(1/2)n but am not able to find the general formula for the sum of this sequence. apparently 1/mn+! doesn't work either.

 

[Mark44]

Homework Statement

what is the general formula for the sequence (1/1*3+1/3*6+1/6*10+1/10*15...)

This is what you wrote:
$$ \frac{1}{1} * 3 +\frac{1}{3} * 6 + \frac{1}{6} * 10 + \frac{1}{10} * 15 ...$$
If that's not what you meant, then use parentheses to clarify it.

Homework Equations

i used the equation n/mn+1 but am not able to use it for this sequence

The Attempt at a Solution

I found the sequence of the denominators which is (1/2)n^2+(1/2)n but am not able to find the general formula for the sum of this sequence. apparently 1/mn+! doesn't work either.

What does 1/mn+? mean?

 

[HallsofIvy]

Assuming that you mean "1/(1*3)+ 1/(3*6)+ 1/(6*10)+..., I would look at the two factors separately. The first factor in each product is 1, 3, 6, 10. The "first difference" is 3-1= 2, 6- 3= 3, and 10- 6= 4. The "second difference" would be 3- 2= 1 and 4- 3= 1, a constant, so we can write this as a quadratic in n: $an^2+ bn+ c$. When n= 1, we have a+ b+ c= 1, when n= 2, we have 4a+ 2b+ c= 3, and when n= 3, we have 9a+ 3b+ c= 6. Solve those three equations for a, b, and c. (And then check that 16a+ 4b+ c= 10.)

Of course, the second factor is just the first factor of the next term in the sum. So if we call the (quadratic) formula for the first factor "A(n)", the second factor is "A(n+1)" and nth term in the sum is 1/(A(n)A(n+1)).

 

[Ray Vickson]

Homework Statement

what is the general formula for the sequence (1/1*3+1/3*6+1/6*10+1/10*15...)

Homework Equations

i used the equation n/mn+1 but am not able to use it for this sequence

The Attempt at a Solution

I found the sequence of the denominators which is (1/2)n^2+(1/2)n but am not able to find the general formula for the sum of this sequence. apparently 1/mn+! doesn't work either.

If you mean 1/(1*3) + 1/(3*6) + 1/(6*10) + ..., you will not be able to find a nice closed-form expression for the finite sum in terms of elementary functions. Maple gets the sum of 1/[f(n)*f(n+1)], for n = 1 to N (and for your f(n)) in terms of the first derivative of the so-called *di-gamma* function, as well as $\pi^2$ and other things. The infinite sum (where N = ∞) is relatively simple: it equals $7 - (2/3) \pi^2$.

 

[vishnu manoj]

sorry about that, i am new to the forum but the sequence is 1/(1*3) + 1/(3*6)+1/(6*10).

 

[vishnu manoj]

Assuming that you mean "1/(1*3)+ 1/(3*6)+ 1/(6*10)+..., I would look at the two factors separately. The first factor in each product is 1, 3, 6, 10. The "first difference" is 3-1= 2, 6- 3= 3, and 10- 6= 4. The "second difference" would be 3- 2= 1 and 4- 3= 1, a constant, so we can write this as a quadratic in n: $an^2+ bn+ c$. When n= 1, we have a+ b+ c= 1, when n= 2, we have 4a+ 2b+ c= 3, and when n= 3, we have 9a+ 3b+ c= 6. Solve those three equations for a, b, and c. (And then check that 16a+ 4b+ c= 10.)

Of course, the second factor is just the first factor of the next term in the sum. So if we call the (quadratic) formula for the first factor "A(n)", the second factor is "A(n+1)" and nth term in the sum is 1/(A(n)A(n+1)).

I did exactly that for the denominator of the sequence but my question is how can you find the sum of all the terms in the sequence with a formula. The quadratic sequence in the denominator really complicates the sum of the sequence.

 

[vishnu manoj]

Thanks Ray Vickson but isn't there a way to find the finite sum of the sequence and if so could you present it ?
It would be really helpful, thanks.

 

[sankalpmittal]

Thanks Ray Vickson but isn't there a way to find the finite sum of the sequence and if so could you present it ?
It would be really helpful, thanks.

If I were given that problem, I would have used method of differences :
Using method of differences :

(i) Find general term of series, say T_n:

1,3,6,10,15,...

(ii) Find general term of series, say T_n':

3,6,10,15,21,...

(iii) Then You get general term of overall series as,

1/(T_n*T_n')

(iv) Then you can sum it up using summation... Then you have to impose limit also as number of terms tend to infinity.

Or if you do not like it, follow HallsOfIvy's method. Then break general term into partial fraction, and sum them up.

 

[Ray Vickson]

If I were given that problem, I would have used method of differences :
Using method of differences :

(i) Find general term of series, say T_n:

1,3,6,10,15,...

(ii) Find general term of series, say T_n':

3,6,10,15,21,...

(iii) Then You get general term of overall series as,

1/(T_n*T_n')

(iv) Then you can sum it up using summation... Then you have to impose limit also as number of terms tend to infinity.

Or if you do not like it, follow HallsOfIvy's method. Then break general term into partial fraction, and sum them up.

That is what Maple did, and it then applied the formulas:
$$\sum_{n=1}^N \frac{1}{n} = \gamma + \Psi(N+1)\\ \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6} - \Psi(1,N+1),$$
where $\Psi(x)$ is the digamma function, $\Psi(1,x)$ is the first derivative of $\Psi(x)$ and $\gamma$ is Euler's constant.

 

[vishnu manoj]

thanks a lot. It really helped