Prove that sequence is monotonic

[Siune]

Homework Statement

Need to prove that the following sequence is monotonic ( decreasing ).

\frac{1}{n^2}+\frac{(-1)^{n}}{3^n}

Homework Equations

-

The Attempt at a Solution

I have idea how to prove that the sequence is decreasing that is:

a_{n+1} - a_{n} ≤ 0
but in this case, I can't get the inequality proved.

\frac{1}{(n+1)^2}+\frac{(-1)^{n+1}}{3^{n+1}} - \big( \frac{1}{n^2}+\frac{(-1)^{n}}{3^n} \big) ≤ 0

I end up in a mess, that looks like;

\frac{3^{n+1}(-2n-1)-2n^{2}(n+1)^2(-1)^n}{n^2((n+1)^2)3^{n+1}}

Could get any hint which way I should start to reform the equation?

Also I'm interested in other thing. If I need to show that for example this sequence is bounded, how I can "see" what the limit "might " be.
In every material I see, that they check the first few terms and make "good guess"?

For example we have sequence a_0 = 0, a_{n+1} = (1/2)x_n +5. We can see when calculating the first we values, that
it seems to converge to x = 10. But if the sequence is more complicated?

 

[ehild]

The 3th terms is 0.074074, the 4th term is 0.07484, greater. The sequence is not monotonous.

ehild

 

[Siune]

Ah, stupid me. Had calculated the first terms totally wrong.. Well, that could have saved some time and not bang my head against the wall! :D

Moderators can lock this up, sorry for inconvience.

 

[Ray Vickson]

Homework Statement

Need to prove that the following sequence is monotonic ( decreasing ).

\frac{1}{n^2}+\frac{(-1)^{n}}{3^n}

Homework Equations

-

The Attempt at a Solution

I have idea how to prove that the sequence is decreasing that is:

a_{n+1} - a_{n} ≤ 0
but in this case, I can't get the inequality proved.

\frac{1}{(n+1)^2}+\frac{(-1)^{n+1}}{3^{n+1}} - \big( \frac{1}{n^2}+\frac{(-1)^{n}}{3^n} \big) ≤ 0

I end up in a mess, that looks like;

\frac{3^{n+1}(-2n-1)-2n^{2}(n+1)^2(-1)^n}{n^2((n+1)^2)3^{n+1}}

Could get any hint which way I should start to reform the equation?

Also I'm interested in other thing. If I need to show that for example this sequence is bounded, how I can "see" what the limit "might " be.
In every material I see, that they check the first few terms and make "good guess"?

For example we have sequence a_0 = 0, a_{n+1} = (1/2)x_n +5. We can see when calculating the first we values, that
it seems to converge to x = 10. But if the sequence is more complicated?

To answer your second question: if $a_{n+1} = (1/2)a_n + 5$, and IF $a_n$ has a limit $L$, you can calculate $L$ from $L = (1/2)L + 5$. In this case you can show that for any value of $a_0$ a limit does, indeed, exist. As for more complicated examples, sometimes the same type of 'trick' works (not always, but sometimes).