Finding an oscillating sequence that diverges and whose limit is zero.

4zimuth
Messages
1
Reaction score
0

Homework Statement



Hi, I need to find an oscillating sequence whose limit of the differences as n approaches infinity is zero but the sequence itself is diverging.

Homework Equations



None.

The Attempt at a Solution



My initial guess was:

\frac{sin(ln(n))}{ln(n)}
 
Physics news on Phys.org
What do you mean by an "oscillating" sequence?
 
Hi!

If I understand correctly, then
sin\left(\sqrt{n}\right)
is what you are looking for. The proof is based on the idea sin(a)-sin(b) < a-b and sqrt(n+1)-sqrt(n) converges to 0.
I originally bumped this post, because I need help with a similar problem, where not
an+1-an converges to 0, but an+1/an converges to 1.
Does anybody know a good example for the latter? So the problem again:
We need a sequence, that:
1) oscillates (oscillation is when it divergates, but neither to infinity nor negative infinity, eg. -1 +1 -1 +1 ... ; +1 -2 +3 -4 +5 -6 ... ;sin(n) ; etc.)
2) an+1/an converges to 1
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top