Does a Monotonic, Decreasing Series Always Converge to 0?

talolard
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Hey guys,

Homework Statement


Prove: If for every n a_{n}>0 and \frac{a_{n+1}}{a_{n}}<1 then the series lim_{n->\infty} a_{n}<0

The Attempt at a Solution


We know that a_{n} is lowerly bounded by 0 and upwardly bounded by a_{1}. we also know that it is monotonic and decreasing and so congerges. But how do I show that it converges to 0. What is to stop it from converging to, say, .5?
Thanks
Tal
 
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When you say series, are you referring to a summation or a sequence?

If an < 0 for all n, then all your terms are negative. So apply this fact to an+1/an < 1
 
Sorry, I made a typo. that was an>0.
 
and I am referring to a sequence, not a summation. Pardon me, english is not my native language.
 
Ahh, I misread the question. it was prove or disprove. I found a counter example.
Thanks anyway.
Tal
 
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...

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