Test Convergence/Divergence of Series: 7*\sum_{n=1}^{\infty} \frac{1}{n^{8n}}

[tangibleLime]

Homework Statement

Test the series for convergence or divergence.

Homework Equations

The Attempt at a Solution

I started playing around with it, trying to write it in different ways, and got it down to something that *looked* like a p-series... but there's an 'n' in the exponent. Does this still make the p-series test valid?

$$\sum_{n=1}^{\infty} \frac{(7n+1)^{n}}{n^{9n}}$$

I removed the one because it seems like it would make no difference as n -> infinity.

$$\sum_{n=1}^{\infty} \frac{(7n)^{n}}{n^{9n}}$$

$$\sum_{n=1}^{\infty} \frac{7}{n^{8n}}$$

Factoring out the 7...

$$7 * \sum_{1}^{\infty} \frac{1}{n^{8n}}$$

I figure since n is always positive and greater than 0, the power 8n will follow the same standards. And 1/n^p is a p-series, and the p in this case is 8n. But is this still valid since there's an 'n' in the power?

 

[JG89]

$$0 < \frac{1}{n^{8n}} < \frac{1}{n^8}$$, right?

 

[PAllen]

Just use the comparison test against a simple power series.

 

[LCKurtz]

Homework Statement

Test the series for convergence or divergence.

Homework Equations

The Attempt at a Solution

I started playing around with it, trying to write it in different ways, and got it down to something that *looked* like a p-series... but there's an 'n' in the exponent. Does this still make the p-series test valid?

$$\sum_{n=1}^{\infty} \frac{(7n+1)^{n}}{n^{9n}}$$

I removed the one because it seems like it would make no difference as n -> infinity.

$$\sum_{n=1}^{\infty} \frac{(7n)^{n}}{n^{9n}}$$

$$\sum_{n=1}^{\infty} \frac{7}{n^{8n}}$$

Factoring out the 7...

$$7 * \sum_{1}^{\infty} \frac{1}{n^{8n}}$$

I figure since n is always positive and greater than 0, the power 8n will follow the same standards. And 1/n^p is a p-series, and the p in this case is 8n. But is this still valid since there's an 'n' in the power?

So, by your heuristic argument, your series "seems" like

$$\sum_{1}^{\infty} \frac{7}{n^{8n}}$$

That's a good intuitive approach, but to make it rigorous you need to apply the general comparison test for positive term series:

$${\hbox If }\lim_{n\rightarrow \infty}\frac{a_n}{b_n} = L$$

where L isn't 0 or ∞ then the series ∑a_n and ∑b_n converge or diverge together. Then you can get your last series to converge by a standard comparison test.

 

[PAllen]

So, by your heuristic argument, your series "seems" like

$$\sum_{1}^{\infty} \frac{7}{n^{8n}}$$

That's a good intuitive approach, but to make it rigorous you need to apply the general comparison test for positive term series:

$${\hbox If }\lim_{n\rightarrow \infty}\frac{a_n}{b_n} = L$$

where L isn't 0 or ∞ then the series ∑a_n and ∑b_n converge or diverge together. Then you can get your last series to converge by a standard comparison test.

I always thought the comparison test was a little different, and need not involve limits, explicitly. Specifically,

If a and b are two non-negative sequences, and there exists k such that for every n > k, a sub n >= b sub n, then, if sum (a) converges then sum (b) converges.

 

[LCKurtz]

I always thought the comparison test was a little different, and need not involve limits, explicitly. Spefically,

If a and b are two non-negative sequences, and there exists k such that for every n > k, a sub n >= b sub n, then, if sum (a) converges then sum (b) converges.

Yes, that is the ordinary comparison test. But say you have, for example, the series

$$\sum_{n=1}^\infty \frac {n+5}{n^3}$$

You might like to compare it with the known convergent p series

$$\sum_{n=1}^\infty \frac {1}{n^2}$$

But you can't, because your original series is greater term by term than the p series. But your intuition tells you that the 5 shouldn't really make any difference, which is correct. It is this sort of problem that the general comparison test handles. See if your calculus book doesn't have some examples.

 

[PAllen]

Yes, that is the ordinary comparison test. But say you have, for example, the series

$$\sum_{n=1}^\infty \frac {n+5}{n^3}$$

You might like to compare it with the known convergent p series

$$\sum_{n=1}^\infty \frac {1}{n^2}$$

But you can't, because your original series is greater term by term than the p series. But your intuition tells you that the 5 shouldn't really make any difference, which is correct. It is this sort of problem that the general comparison test handles. See if your calculus book doesn't have some examples.

You still don't need anything more than than the simple comparison test. If series S converges, then series k S converges for any k.

I was last in school 37 years ago. I just play with math and physics as a hobby, unrelated to my current career.

 

[LCKurtz]

Yes, that is the ordinary comparison test. But say you have, for example, the series

$$\sum_{n=1}^\infty \frac {n+5}{n^3}$$

You might like to compare it with the known convergent p series

$$\sum_{n=1}^\infty \frac {1}{n^2}$$

But you can't, because your original series is greater term by term than the p series. But your intuition tells you that the 5 shouldn't really make any difference, which is correct. It is this sort of problem that the general comparison test handles. See if your calculus book doesn't have some examples.

You still don't need anything more than than the simple comparison test. If series S converges, then series k S converges for any k.

I was last in school 37 years ago. I just play with math and physics as a hobby, unrelated to my current career.

Of course the example I gave can be done other ways but I was making the point about when the generalized comparison test is handy. Here's another

$$\sum_{n=1}^\infty \frac {\sqrt{n^2+5}}{n^3-9}$$

 

[PAllen]

Of course the example I gave can be done other ways but I was making the point about when the generalized comparison test is handy. Here's another

$$\sum_{n=1}^\infty \frac {\sqrt{n^2+5}}{n^3-9}$$

Of course I agree the limit approach is useful. I am not sure I believe either limit or comparison are more general. I think if I thought about it for a while, I could prove either from the other.