Series Convergence: Exploring the Limit and Root Test Methods

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Homework Statement


Determine whether the series converges or diverges.


\sum_{n=1}^{\infty}\left(e-\left(1+\frac{1}{n}\right)^n\right)^p where p is a parameter[/tex]


The Attempt at a Solution



\lim_{n\rightarrow\infty}e-\left(1+\frac{1}{n}\right)^n=0

so by using Root Test i decided that

\limsup_{n\rightarrow\infty}\sqrt[n]{\left(e-\left(1+\frac{1}{n}\right)^n\right)^p}<1
Which gives that series converges
 
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What if p=0, for example?
 
Oh,I see.Limit is 1.I must try another way.
 
\sum_{n=1}^{\infty}\left(e-\left(1+\frac{1}{n}\right)^n\right)^p=\sum_{n=1}^{\infty}\left(e-e^{n\ln \left(1+\frac{1}{n}\right)}\right)^p=\sum_{n=1}^{\infty}\left(e-e^{n\left(\frac{1}{n}-\frac{1}{2n^2}+O(\frac{1}{n^3})\right)}\right)^p=\sum_{n=1}^{\infty}\left(e-e^{1-\frac{1}{2n}+O(\frac{1}{n^2})}\right)^p

=\sum_{n=1}^{\infty}e^p\left(1-e^{-\frac{1}{2n}+O(\frac{1}{n^2})}\right)^p=\sum_{n=1}^{\infty}e^p\left(1-(1-\frac{1}{2n}+O(\frac{1}{n^2}))\right)^p

=\sum_{n=1}^{\infty}e^p\left(\frac{1}{2n}+O(\frac{1}{n^2})\right)^p

Is it converges for all p<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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