Does the Series 1/(3n-2)(3n+1) Converge and What is Its Sum?

[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.

 

[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.

 

[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.

 

[adriank]

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}<br /> = \ln (n - 1) - 2 \ln n + \ln (n + 1)

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

 

[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.

 

[adriank]

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.

 

[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?

 

[adriank]

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.

 

[squenshl]

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

 

[adriank]

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*}<br /> \sum_{n=0}^{N} \frac{1}{\sqrt{n} + \sqrt{n+1}}<br /> &amp;= \sum_{n=0}^{N-1} \frac{1}{\sqrt{n} + \sqrt{n+1}} + \frac{1}{\sqrt{N} + \sqrt{N+1}} \\<br /> &amp;= \sqrt{N} + \frac{1}{\sqrt{N} + \sqrt{N+1}},<br /> \end{align*}

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

 

[squenshl]

It goes from n=1 to infinity.

 

[adriank]

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.

 

[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))

 

[squenshl]

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

 

[adriank]

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))

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