Sum convergence - is my approach flawless?

[twoflower]

I have to find out, whether this sum converges:

$$\sum_{n = 2}^{+ \infty} n^{\alpha} \left( \log (n+1) - \log n \right)^4$$

So I rewrote it:

$$\sum_{n = 2}^{+ \infty} n^{\alpha} \left( \log \left(1 + \frac{1}{n} \right) \right)^4 =$$

$$\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4$$

Now I can see that the expression with log goes to 1, so the sum gets reduced to

$$\sum_{n = 2}^{+ \infty} \frac {1}{n^{4 - \alpha}}$$

and the result is obviously $\alpha \in \left(0, 3\right)$.

But, I don't know how exactly should I justify my decision. I'm not talking about the very last step now, but the one in which I threw away the logarithm because of the limit 1. We have to give reasons for each non-trivial step we do, so I think I should write some theorem or rule under that step...

Thank you.

 

[Galileo]

If you're saying that:

$$\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4=\sum_{n = 2}^{+ \infty} \frac{1}{n^{4-\alpha}}$$
..then no.
But you can use it to compare the original series it with the right hand side.

If you can show
$$\left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4<1$$
and approaches 1, then you're home free. That means the original series converges too.

 

[twoflower]

If you're saying that:

$$\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4=\sum_{n = 2}^{+ \infty} \frac{1}{n^{4-\alpha}}$$
..then no.
But you can use it to compare the original series it with the right hand side.

If you can show
$$\left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4<1$$
and approaches 1, then you're home free. That means the original series converges too.

Yes, I see what you mean. But...how should I prove this:

If you can show
$$\left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4<1$$

I only know the limit is 1, nothing more...

 

[learningphysics]

Yes, I see what you mean. But...how should I prove this:

If you can show
$$\left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4<1$$

I only know the limit is 1, nothing more...

Use the limit comparison test. You know that $$n^{\alpha-4}$$ converges for $$\alpha <= 2$$.

Compare your given series

$$\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4$$

to the series $$n^{\alpha-4}$$, and calculate the limit... which you've already discovered is 1 (this is essentially what you did anway). Since this is finite and greater than zero, your series must converge.

This also shows that when $$n^{\alpha-4}$$ diverges ( $$\alpha > 2$$, your series also diverges.

 

[twoflower]

Use the limit comparison test. You know that $$n^{\alpha-4}$$ converges for $$\alpha <= 2$$.

Compare your given series

$$\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4$$

to the series $$n^{\alpha-4}$$, and calculate the limit... which you've already discovered is 1 (this is essentially what you did anway). Since this is finite and greater than zero, your series must converge.

This also shows that when $$n^{\alpha-4}$$ diverges ( $$\alpha > 2$$, your series also diverges.

Now it's clear, thank you learningphysics and Galileo. Limit comparsion test is the clue.

 

[twoflower]

Use the limit comparison test. You know that $$n^{\alpha-4}$$ converges for $$\alpha <= 2$$.

Compare your given series

$$\sum_{n = 2}^{+ \infty} n^{\alpha - 4} \left[ \frac {\log \left(1 + \frac{1}{n} \right)}{\frac{1}{n}} \right]^4$$

to the series $$n^{\alpha-4}$$, and calculate the limit... which you've already discovered is 1 (this is essentially what you did anway). Since this is finite and greater than zero, your series must converge.

This also shows that when $$n^{\alpha-4}$$ diverges ( $$\alpha > 2$$, your series also diverges.

I thought it was clear to me, but some time ago I had to solve similar problem and I realized that comparing my original sum to some other sum only tells me, for which $\alpha$ it converges surely, but it doesn't tell me all $\alpha$ values for which it converges. I hope I expressed it in an understandable way...