# Does this series converge

1. Apr 6, 2009

### squenshl

I have a problem.
Does the series 1/(3n-2)(3n+1) converge or diverge. If it converges find its sum.
How would I do this.

2. Apr 6, 2009

### squenshl

Also, does the series ln(1-1/n^2) converge or diverge. If so calculate the sum.
I know to first calculate the partial sum but i don't know how to do this. Someone please help.

3. Apr 6, 2009

### AUMathTutor

The first series converges for sure. To see this, simply note that it is asymptotically less than 1/n^2, a convergent series.

If the second one starts at n=1, then I'm sure it doesn't converge. Plugging in n=1, you get that the first term is negative infinity.

To evaluate the first series... well, that's a good question.

4. Apr 7, 2009

Observe that

$$\frac{1}{(3n-2)(3n+1)} = \frac13 \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right)$$

so you get a telescoping series. Similarly,

$$\ln \left( 1 - \frac{1}{n^2} \right) = \ln \frac{(n - 1)(n + 1)}{n^2} = \ln (n - 1) - 2 \ln n + \ln (n + 1)$$

which also collapses on summation. (Of course, you must start at n = 2.)

Last edited: Apr 7, 2009
5. Apr 10, 2009

### squenshl

One more problem.
How do I calculate the Nth partial sum to the series 1/(sqrt(n)+sqrt(n+1)) and hence determine if this series converges.
And determine if the series sin^2(1/n) diverges or converges.
Someones help would be appreciated.
I'm not to sure how to start either of them.

6. Apr 10, 2009

For the first one, you should find that the first few partial sums

$$\sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}}$$

are $$1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}$$. Can you make a conjecture from this and prove it (by induction)?

For the second one, compare with 1/n2.

Last edited: Apr 10, 2009
7. Apr 12, 2009

### squenshl

Why do I prove it by induction?
And I'm still not sure how to find the Nth sum. What is the meaning of the Nth sum?

8. Apr 12, 2009

If you have a series
$$\sum_{n=0}^{\infty} a_n,$$
the Nth partial sum is
$$\sum_{n=0}^N a_n.$$

As for proving that, I'm just suggesting a way to prove it, that's probably the easiest.

9. Apr 12, 2009

### squenshl

Okay, I see now. Thanx.
So how would I apply it to the example 1/(sqrt(n)+sqrt(n+1))?

10. Apr 12, 2009

Well, as I suggested, your conjecture should be that

$$\sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{N + 1}.$$

You would prove that by induction on N. (Have you done induction proofs before? Any good analysis course should teach it.) Clearly it's true if N = 0; now suppose it's true for N - 1, that is, suppose that

$$\sum_{n=0}^{N-1} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{(N - 1) + 1} = \sqrt{N}.$$

Then,

\begin{align*} \sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}} &= \sum_{n=0}^{N-1} \frac{1}{\sqrt{n} + \sqrt{n+1}} + \frac{1}{\sqrt{N} + \sqrt{N+1}} \\ &= \sqrt{N} + \frac{1}{\sqrt{N} + \sqrt{N+1}}, \end{align*}

and all you have to do now is prove that this equals $$\sqrt{N + 1}$$.

11. Apr 12, 2009

### squenshl

It goes from n=1 to infinity.

12. Apr 12, 2009

Then just subtract the n = 0 term from everything, so

$$\sum_{n=1}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{N + 1} - 1.$$

13. Apr 12, 2009

### squenshl

Thanks heaps for that.
How would I prove that this converges then.
Do I do a comparison test with sqrt(N+1)-1 and 1/(sqrt(n)+sqrt(n+1))

14. Apr 14, 2009

### squenshl

To see if the series sin^2(1/n) converges or diverges do I do a comparison test with 1/n^2.
Thanks.

15. Apr 15, 2009

No; you'd use the definition of convergence of a series.