# Infinite Series Test (Ratio Test get 1)

## Homework Statement

Test if the infinite series converge or diverge.

## Homework Equations

$$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}$$

## The Attempt at a Solution

I tried Ratio test:
$$a_{n+1} = \frac{4n+7}{(n+1)(n+2)(n+3)}$$
$$a_{n} = \frac{4n+3}{n(n+1)(n+2)}$$
$$\left|\frac{a_{n+1}}{a_{n}}\right| = \frac{4n+7}{(n+1)(n+2)(n+3)} \times \frac{n(n+1)(n+2)}{4n+3} = \frac{n(4+7n)}{(n+3)(4n+3)} = \frac{4n^{2}+7n}{4n^{2}+15n+9}$$
$$lim_{n\rightarrow\infty} \left|\frac{a_{n+1}}{a_{n}}\right| = lim_{n\rightarrow\infty} \frac{4+\frac{7}{n}}{4+\frac{15}{n}+\frac{9}{n^{2}}} = \frac{4}{4} = 1$$
The answer is inconclusive, and I can't seem to think of any other test yet.
Anyone can help me with this?
I will much appreciate it. Thanks!

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
Think about what the terms look like for large n. Try a comparison test.

Think about what the terms look like for large n. Try a comparison test.
This is what I done:
$$n^{3} > n(n+1)(n+1)$$
$$\frac{4n+3}{n^{3}} < \frac{4n+2}{n(n+1)(n+2)}$$
$$\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}$$
Both converge

Therefore, $$\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}$$ converges.

Is that correct?

Last edited:
How is n^3 > n^3 + 2n^2 + n?

How is n^3 > n^3 + 2n^2 + n?
Isn't this the same with the $$n^{3} > n(n+1)(n+1)$$ above?

Mark44
Mentor
Isn't this the same with the $$n^{3} > n(n+1)(n+1)$$ above?
Yes, so why do you think that n3 > n(n + 1)(n + 2)?

Yes, so why do you think that n3 > n(n + 1)(n + 2)?
Opps, sorry,
found the careless mistake.

should be
n(n+1)(n+2) > n3

so
$$\frac{4n+3}{n(n+1)(n+2)} < \frac{4n+3}{n^{3}}$$
Thanks for pointing me out

$$\frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}}$$
Both converge

Therefore, $$\sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)}$$ converges.

(use the term by term comparison test)

$$\frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}$$

(use the term by term comparison test)

$$\frac{4n+3}{n(n+1)(n+2)}\leq \frac{7n}{n^{3}}= 7\frac{1}{n^{2}}$$
How do you get the 7n?

Mark44
Mentor
4n + 3 <= 7n for all n >= 1

4n + 3 <= 7n for all n >= 1
Got it!
Thanks!

So, the final answer should look like this:
$$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}$$

n(n+1)(n+2) $$\geq$$ n3
$$\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}$$

4n+3 $$\leq$$ 7n , for all n $$\geq$$ 1

$$\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}$$

$$\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
Converge P-series (p > 1)

According to Comparison test,
since $$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
$$and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges$$,
$$therefore \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.$$

Dick
Homework Helper
So, the final answer should look like this:
$$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)}$$

n(n+1)(n+2) $$\geq$$ n3
$$\frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}}$$

4n+3 $$\leq$$ 7n , for all n $$\geq$$ 1

$$\frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}}$$

$$\sum_{n=1}^{\infty}\frac{7n}{n^{3}} = \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
Converge P-series (p > 1)

According to Comparison test,
since $$\sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} \leq \sum_{n=1}^{\infty}\frac{7}{n^{2}}$$
$$and \sum_{n=1}^{\infty}\frac{7}{n^{2}} converges$$,
$$therefore \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} converges.$$
Very nice.

Very nice.
Thanks,
and thanks for everyone that helps. 