Finding the supremum of the following set I am stuck on the following problem:  

Let $P=\{x \in \Bbb R: x\ge 0,\sum_{n=1}^{\infty}x^{\sqrt n}< \infty\}$.Then what is the supremum of $P$?  

Can someone help me out by providing some explanation? Thanks in advance.
 A: What you are looking for is the supremum of the values for which your series converges (if I got it right). Now you only got to proceed with method. 
First of all you can observe that for $x \geq 1$ the succession does not go to zero for $n \rightarrow \infty$, so the series obviously diverges. For $x=0$ the series converges. Now let us consider $x \in (0,1)$. We know $f(a)=x^{ \sqrt{a}}$ is monotonically decreasing (and it's limit is $0$ for $a \rightarrow \infty$) and positive. So we can use the indefinite integral, knowing that $$\sum_2^n f(k) \leq \int_1^n f(a) d a \leq \sum_1^n f(k).$$Now we need to evaluate the integral: $$\int_1^n x^{\sqrt{a}}da= \frac{1}{\ln(x)}\int_1^{\sqrt{n}}2y*x^y \ln(x)dy=$$Now we use integration by parts: $$=\frac{2y x^y}{\ln(x)} \left. \right|_1^{\sqrt{n}}-\frac{2}{\ln^2(x)}\int_1^{\sqrt{n}}x^y \ln(x)dy=...=\frac{2x^{\sqrt{n}}}{\ln^2(x)}(\sqrt{n}\ln(x)-1)+c,$$ where $c$ is the value of the primitive in $1$. 
Now if you evaluate the limit on $n \rightarrow \infty$, you find $c$, which proves that the integral converges. Now this means that $P=[0,1)$, which has it's sup in $1.$
