Determine if the sum Converges or Diverges

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Homework Statement



dve9e.jpg


The Attempt at a Solution



1. I have no idea. I know that the summation of the series converges.


2. I think it would diverge because the limit of the function does not equal zero.


3. I have tried the ratio test and got 1. Can't use the alternating series test because when ignoring signs the function increases.
 
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For 3 what's the limit of (5n-1)/(n+5) and what does that tell you about convergence? For 1 I think they might be asking you to estimate the difference using an integral.
 
2) is correct
3) Think about the function f(x)=\ln x
 
Last edited:
Dick said:
For 3 what's the limit of (5n-1)/(n+5) and what does that tell you about convergence? For 1 I think they might be asking you to estimate the difference using an integral.

The limit of (5n-1)/(n+5) as x approaches infinity equals 5 thus meaning that the series in divergent? I think
 
hunt_mat said:
2) is correct
3) Think about the function f(x)=\ln x
[/QUOTE]<br /> <br /> I&#039;m not sure what f(x)=\ln x[/quote] means. Can you plus explain.
 
McAfee said:
The limit of (5n-1)/(n+5) as x approaches infinity equals 5 thus meaning that the series in divergent? I think

Right. If the nth term of a series doesn't approach 0 then it's always divergent. Now can you write an integral that's greater than the difference in 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...

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