Thread: How can i check if this sum is converges?

1. How can i check if this sum is converges?

Hello everybody,
How can i check if this sum is converges?

2. Re: How can i check if this sum is converges?

Hey MrMath.

Hint - Check the denominator in terms of the power and make a deduction with some kind of p-series.

3. Re: How can i check if this sum is converges?

Originally Posted by chiro
Hey MrMath.

Hint - Check the denominator in terms of the power and make a deduction with some kind of p-series.
Can you explain more?
I tried to do it with convergence test but i didn't success. i will be happy if you can present a full solution.

4. Re: How can i check if this sum is converges?

$\sum^\infty_{n=1} \frac{40n}{(n+3)(n+4)(n+5)}=40\cdot \sum^\infty_{n=1} \frac{n}{(n+3)(n+4)(n+5)}$

Let $\sum a_n, \sum b_n$ be two positive sequences such that $a_n\leq b_n$ for all $n$.

If $\sum b_n$ converges, so does $\sum a_n$. If $\sum a_n$ diverges, so does $\sum b_n$.

$\sum^\infty_{n=1}\frac{n}{(n+3)(n+4)(n+5)} \leq \sum^\infty_{n=1} \frac{n}{n^3} = \sum^\infty_{n=1} \frac{1}{n^2}$

We now have a series of the form $\sum^\infty_{n=1}\frac{1}{n^p}$. Can you take it from here?

-Andy

5. Re: How can i check if this sum is converges?

Originally Posted by abender
$\sum^\infty_{n=1} \frac{40n}{(n+3)(n+4)(n+5)}=40\cdot \sum^\infty_{n=1} \frac{n}{(n+3)(n+4)(n+5)}$

Let $\sum a_n, \sum b_n$ be two positive sequences such that $a_n\leq b_n$ for all $n$.

If $\sum b_n$ converges, so does $\sum a_n$. If $\sum a_n$ diverges, so does $\sum b_n$.

$\sum^\infty_{n=1}\frac{n}{(n+3)(n+4)(n+5)} \leq \sum^\infty_{n=1} \frac{n}{n^3} = \sum^\infty_{n=1} \frac{1}{n^2}$

We now have a series of the form $\sum^\infty_{n=1}\frac{1}{n^p}$. Can you take it from here?

-Andy
Yes. THANK YOU!!

6. Re: How can i check if this sum is converges?

Observe that for $n>0$: $$\large \frac{40n}{(n+3)(n+4)(n+5)}< \frac{40n}{n^3}=\frac{40}{n^2},$$ then:
$$S_N=\sum_{n=1}^{N}\frac{40n}{(n+3)(n+4)(n+5)}< \sum_{n=1}^{N}\frac{40}{n^2},$$ so $S_N$ is an increasing sequence bounded above by $K=\sum_{n=1}^{\infty}\frac{40}{n^2}$ (which is convergent with sum $\pi^2/6$). An increasing sequence bounded above converges...