Convergence Criteria for Series with Exponential and Polynomial Terms

wanchosen
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I have been stuck on this question for a while and have not got much success from class mates...can somebody please help?

Using an appropriate convergence test, find the values of x \in R for which the following series is convergent:

\sumn=1n \frac{1}{e^n * n^x}

So,

Un = \frac{1}{e^n * n^x}

and

U(n+1) = \frac{1}{e^(n+1) * (n+1)^x}

then

U(n+1)/Un = \frac{n^x}{e * (n+1)^x}

If the series is convergent:

\frac{n^x}{(n+1)^x} < e

ln(\frac{n^x}{(n+1)^x}) < 1

x * ln(\frac{n}{n+1}) < 1

n/n+1 is always positive but less than one, so the Ln term is negative,

So

x > \frac{1}{ln(n/n+1)}

But the limit of this as n tends to infininty, is infinity, which doesn't make sense. Can I solve this for x, independent of n?
 
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Your trouble is in the last step. You divide both sides by the log without realizing that as n->+inf, it is equal to zero.

Clearly, and you did all the work mind you, this is true for any x at all.

Reason? |x ln(n/n+1)| < 1 for all x you can think of... since any real number times zero equals zero, and this is less than 1 in absolute value.

Good job! Careful about that algebra, though...
 
Thanks!
 
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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