Are these the best tests for convergence of the following series?

end3r7
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Homework Statement


(a) \displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}}{n^{1 + \frac{1}{n}}}\Bigg)
(b) \displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{e^{\frac{1}{n}}}{n^{2}}\Bigg)
(c) \displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{(-1)^{n}n!}{n^{n}}\Bigg)
(d) \displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{1}{n^{1 + \frac{1}{n}}}\Bigg)
(e) \displaystyle{\sum_{n=1}^{\infty}\Bigg(\frac{ln(x)}{n^{\frac{3}{2}}}\Bigg)
(f) \displaystyle{\sum_{n=1}^{\infty}\Bigg(1 - e^{\frac{-1}{n}}\Bigg)

Homework Equations


The test that we have messed with are:
Telescoping, Geometric, P-Series, Ratio, Root, Simple Comparison, Limit Comparison, Absolute Convergence, Alternating Series, Dirichlet, Integral, Gauss

The Attempt at a Solution



First, I got they all converge:

(a) (d)
I worked 'd' first. I did a limit comparison test with 1/n
\frac{\frac{1}{n}}{\frac{1}{n^{1 + \frac{1}{n}}}} = n^{n}
That limit is 1. And since its absolute value converges (a) converges.

(b) Basic Comparion test wtih <= \frac{e}{n^{2}}

(c) Alternating series test

(e) Integral test

(f) I am not sure which test to apply
 
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(d) is wrong,a dn so is (a), not sure how to prove if they are convergent (if they are).

Is (d) divergent?
 
(a) is convergent by alternating series test.

Did I end up proving (d) divergent or is my proof wrong?

I'm also quite lost on (f) since I can't integrate that function nicely
 
Too many questions at once! d) is divergent. It's the same as 1/(n*n^(1/n)). For n^(1/n) show that the log of that approaches zero. So n^(1/n) approaches 1. So you can do a comparison with say 1/(2n). For f) expand e^(-1/n) in a power series using e^x=1+x+x^2/2!+etc and keep only the terms that matter.
 
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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