simple and uniform convergence of $ \sum ( x+\frac{1}{n})^{n+\frac{x}{n}} $ It's my first post in this forum ! Hello ! (I have a very bad english sorry in advance...) I come with an exercice that I can not solve. I hope you can help me...
Q1 : simple and uniform convergence of $ \sum ( x+\frac{1}{n})^{n+\frac{x}{n}} $
Here my guess and what I have done the convergence is |x|< 1 (that I didn't proved just guessed)
I tried limited development it didn't worked.
Thank you for reading !
 A: One possible solution is to apply the $n$-root test.
To have powers well defined, assume $x\geq0$.
Let $a_n(x)=\Big(x+\frac{1}{n}\Big)^{n+\tfrac{x}{n}}=x^{n}x^{x/n}\Big(1+\frac{1}{xn}\Big)^n\Big(1+\frac{1}{xn}\Big)^{x/n}$. Then
$$
|a_n(x)|^{1/n}=|x||x|^{x/n^2}\Big|1+\frac{1}{xn}\Big|\Big|1+\frac{1}{nx}\Big|^{x/n^2}\xrightarrow{n\rightarrow\infty}|x|
$$
Hence, as you suspected, convergence holds for $0<x<1$ and uniform convergence occurs in compact subsets of $(0,1)$. Some details are left for the OP.
A: For the convergence, using the ratio test, let
$$a_n=\left( x+\frac{1}{n}\right)^{n+\frac{x}{n}}\implies \log(a_n)=\left(n+\frac{x}{n}\right) \log \left(x+\frac{1}{n}\right)$$ Now, using Taylor series for large values of $n$
$$\log(a_n)=n \log (x)+\frac{1}{x}+\frac{2x^3 \log (x)-1}{2x^2n}+O\left(\frac{1}{n^2}\right)$$ Apply it twice and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=\log (x)+\frac{1-2x^3 \log
   (x)}{2x^2n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} } {a_n}=\exp\big(\log(a_{n+1})-\log(a_n) \big)=x+\frac{1-2 x^3 \log (x)}{2 n^2 x}+O\left(\frac{1}{n^3}\right)$$
