1. ## Convergence problem

Test the convergence of the series
$\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+.. .$

2. $\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}$

You can use any number of tests to see if it converges or not(ratio test), but if it converges what does it converges to?.

3. ## which test to use

Which test should i use to evaluate this ..

4. Does the fact that $\frac{2n-1}{n(n+1)(n+2)}< \frac{2}{n^2}$ help at all?

5. Originally Posted by galactus
$\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}$

You can use any number of tests to see if it converges or not(ratio test), but if it converges what does it converges to?.
Now that's a good question!

6. Split that into two series (both of them converge because of direct comparison test) and then:

$\frac{1}{(n+1)(n+2)}=\frac{n!}{(n+2)!}=\frac{\Gamm a (n+1)\Gamma (2)}{\Gamma (n+3)}=\beta (n+1,2).$

Hence, turn that into a beta integral and switch the integral by the series. Use geometric series and solve the remaining integral. Do the same with the other.

In general for $n\ge2,$ $\sum\limits_{k=0}^{\infty }{\frac{1}{(k+1)(k+2)\cdots (k+n)}}=\frac{1}{(n-1)(n-1)!},$ which can be proved in the same fashion when using Beta function.

7. Originally Posted by Krizalid
Split that into two series (both of them converge because of direct comparison test) and then:

$\frac{1}{(n+1)(n+2)}=\frac{n!}{(n+2)!}=\frac{\Gamm a (n+1)\Gamma (2)}{\Gamma (n+3)}=\beta (n+1,2).$

Hence, turn that into a beta integral and switch the integral by the series. Use geometric series and solve the remaining integral. Do the same with the other.

In general for $n\ge2,$ $\sum\limits_{k=0}^{\infty }{\frac{1}{(k+1)(k+2)\cdots (k+n)}}=\frac{1}{(n-1)(n-1)!},$ which can be proved in the same fashion when using Beta function.
There's also a very nice pattern if you decompose the series, i.e.

$\frac{1}{n(n+1) }= \frac{1}{n}-\frac{1}{n+1}$

$\frac{1}{n(n+1) (n+2)}= \frac{1}{2}\left(\frac{1}{n}-\frac{2}{n+1} + \frac{1}{n+2}\right)$

$\frac{1}{n(n+1) (n+2)(n+3)}= \frac{1}{6}\left(\frac{1}{n}-\frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$

$\frac{1}{n(n+1) (n+2)(n+3)(n+4)}= \frac{1}{4!}\left(\frac{1}{n}-\frac{4}{n+1} + \frac{6}{n+2} - \frac{4}{n+3} + \frac{1}{n+4}\right)$

Then one can group the terms as to make it a telescopic series. Just my 2 cents.

8. This is similar to Krizalid's post except I am not sure this is what he had in mind

Consider

$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n~~|x|<1$

I dont feel like justifying interchanging the integral and summation, but just notice that your sum

\begin{aligned}\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}&=\sum_{n=1}^{\infty}\frac{2}{(n+1) (n+2)}-\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\\
&=2\int_0^1\left\{\int\frac{dx}{1-x}\right\}-\int_0^1\left\{\iint\frac{dx}{x(1-x)}\right\}\\
&=\frac{3}{4}\end{aligned}

9. Originally Posted by Mathstud28
This is similar to Krizalid's post except I am not sure this is what he had in mind

Consider

$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n~~|x|<1$

I dont feel like justifying interchanging the integral and summation, but just notice that your sum

\begin{aligned}\sum_{n=1}^{\infty}\frac{2n-1}{n(n+1)(n+2)}&=\sum_{n=1}^{\infty}\frac{2}{(n+1) (n+2)}-\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}\\
&=2\int_0^1\left\{\int\frac{x}{1-x}dx\right\}-\int_0^1\left\{\iint\frac{dx}{1-x}\right\}\\
&=\frac{3}{4}\end{aligned}
But the series doesn't converge at $x = 1$

10. Originally Posted by danny arrigo
But the series doesn't converge at $x = 1$
It is obvious that any integral or derivative of a power series evaluated at its center (in this case 0) is 0. So instead let us consider

\begin{aligned}\iiint\frac{1}{x(1-x)}\bigg|_{x=1}&=\iiint\sum_{n=1}^{\infty}x^{n-1}\bigg|_{x=1}\\
&=\sum_{n=1}^{\infty}\iiint x^{n-1} dx|_{x=1}\\
&=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n(n+1)(n+2)}\b igg|_{x=1}\\
&=\sum_{n=0}^{\infty}\frac{1}{n(n+1)(n+2)}\end{ali gned}

Is that what you mean?

11. Originally Posted by Mathstud28
It is obvious that any integral or derivative of a power series evaluated at its center (in this case 0) is 0. So instead let us consider

\begin{aligned}\iiint\frac{1}{x(1-x)}\bigg|_{x=1}&=\iiint\sum_{n=1}^{\infty}x^{n-1}\bigg|_{x=1}\\
&=\sum_{n=1}^{\infty}\iiint x^{n-1} dx|_{x=1}\\
&=\sum_{n=0}^{\infty}\frac{x^{n+2}}{n(n+1)(n+2)}\b igg|_{x=1}\\
&=\sum_{n=0}^{\infty}\frac{1}{n(n+1)(n+2)}\end{ali gned}

Is that what you mean?
What I mean is that you're using
$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ but it doesn't converge at $x=1$

12. Originally Posted by danny arrigo
What I mean is that you're using
$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ but it doesn't converge at $x=1$
So? The endpoints of a radius of convergence are subject to change when integrating.

13. Originally Posted by Mathstud28
So? The endpoints of a radius of convergence are subject to change when integrating.
Okay - after two integrations it does.

14. ## What is the right answer

Can any body tell me the right answer

15. Originally Posted by varunnayudu
Can any body tell me the right answer
Do you mean the actual value? Mathcad gives $\frac{3}{4}$

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