Sum convergence - is my approach flawless?

[twoflower]

I have to find out, whether this sum converges:

<br /> \sum_{n = 2}^{+ \infty} n^{\alpha} \left( \log (n+1) - \log n \right)^4<br />

So I rewrote it:

<br /> \sum_{n = 2}^{+ \infty} n^{\alpha} \left( \log \left(1 + \frac{1}{n} \right) \right)^4 = <br />

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

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

<br /> \sum_{n = 2}^{+ \infty} \frac {1}{n^{4 - \alpha}}<br />

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&lt;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&lt;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&lt;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&lt;1

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

Use the limit comparison test. You know that n^{\alpha-4} converges for \alpha &lt;= 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 &gt; 2, your series also diverges.

 

[twoflower]

Use the limit comparison test. You know that n^{\alpha-4} converges for \alpha &lt;= 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 &gt; 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 &lt;= 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 &gt; 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...